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2.4: Solving Linear Equations- Part II


Learning Objectives

  • Solve general linear equations.
  • Identify and solve conditional equations, identities, and contradictions.
  • Clear decimals and fractions from equations.
  • Solve literal equations or formulas for a given variable.

Combining Like Terms and Simplifying

Linear equations typically are not given in standard form, so solving them requires additional steps. These additional steps include simplifying expressions on each side of the equal sign using the order of operations.

Same-Side Like Terms

We will often encounter linear equations where the expressions on each side of the equal sign can be simplified. Typically, this involves combining same-side like terms. If this is the case, then it is best to simplify each side first before solving.

Example (PageIndex{1})

Solve:

(−4a+2−a=3−2).

Solution:

First, combine the like terms on each side of the equal sign.

Answer:

The solution is (frac{1}{5}).

Opposite-Side Like Terms

Given a linear equation in the form (ax+b=cx+d), we begin by combining like terms on opposite sides of the equal sign. To combine opposite-side like terms, use the addition or subtraction property of equality to effectively “move terms” from one side to the other so that they can be combined.

Example (PageIndex{2})

Solve:

(−2y−3=5y+11).

Solution:

To “move” the term (5y) to the left side, subtract it on both sides.

(egin{aligned} -2y-3&=5y+11 -2y-3color{Cerulean}{-5y}&=5y+11color{Cerulean}{-5y} &color{Cerulean}{Subtract:5y:from:both:sides.} -7y-3&=11 end{aligned})

From here, solve using the techniques developed previously.

Always check to see that the solution is correct by substituting the solution back into the original equation and simplifying to see if you obtain a true statement.

(egin{aligned} -2y-3&=5y+11 -2(color{OliveGreen}{-2}color{black}{)-3}&=5(color{OliveGreen}{-2}color{black}{)+1} 4-3&=-10+11 1&=1 quadcolor{Cerulean}{checkmark} end{aligned})

Answer:

The solution is (-2).

General Guidelines for Solving Linear Equations

When solving linear equations, the goal is to determine what value, if any, will produce a true statement when substituted in the original equation. Do this by isolating the variable using the following steps:

Step 1: Simplify both sides of the equation using the order of operations and combine all same-side like terms.

Step 2: Use the appropriate properties of equality to combine opposite-side like terms with the variable term on one side of the equation and the constant term on the other.

Step 3: Divide or multiply as needed to isolate the variable.

Step 4: Check to see if the answer solves the original equation.

Example (PageIndex{3})

Solve:

(-frac{1}{2}(10y-2)+3=14)

Solution:

Simplify the linear expression on the left side before solving.

To check,

(egin{aligned} -frac{1}{2}(10(color{OliveGreen}{-2}color{black}{)-2)+3}&=14 -frac{1}{2}(-20-2)+3&=14 -frac{1}{2}(-22)+3&=14 11+3&=14 14&=14quadcolor{Cerulean}{checkmark} end{aligned})

Answer:

The solution is (-2).

Example (PageIndex{4})

Solve:

(5(3x+2)−2=−2(1−7x)).

Solution:

First, simplify the expressions on both sides of the equal sign.

Answer:

The solution is (−10). The check is left as an exercise.

Exercise (PageIndex{1})

Solve:

(6−3(4x−1)=4x−7).

Answer

(x=1)

Conditional Equations, Identities, and Contradictions

There are three different types of equations. Up to this point, we have been solving conditional equations. These are equations that are true for particular values. An identity is an equation that is true for all possible values of the variable. For example,

(x=xquadcolor{Cerulean}{Identity})

has a solution set consisting of all real numbers, (R). A contradiction is an equation that is never true and thus has no solutions. For example,

(x+1=xquadcolor{Cerulean}{Contradiction})

has no solution. We use the empty set, (∅), to indicate that there are no solutions.

If the end result of solving an equation is a true statement, like (0 = 0), then the equation is an identity and any real number is a solution. If solving results in a false statement, like (0 = 1), then the equation is a contradiction and there is no solution.

Example (PageIndex{5})

Solve:

(4(x+5)+6=2(2x+3)).

Solution:

(egin{aligned} 4(x+5)+6&=2(2x+3)&color{Cerulean}{Distribute.} 4xcolor{OliveGreen}{+20+6}&=4x+6&color{Cerulean}{Combine:same-side:like:terms.} 4x+26&=4x+6 &color{Cerulean}{Combine:opposite-side:like:terms.} 4x+26color{Cerulean}{-4x}&=4x+6color{Cerulean}{-4x} 26&=6quadcolor{red}{x} &color{Cerulean}{False} end{aligned})

Answer:

(∅). Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.

Example (PageIndex{6})

Solve:

(3(3y+5)+5=10(y+2)−y).

Solution:

(egin{aligned} 3(3y+5)+5&=10(y+2)-y &color{Cerulean}{Distribute.} 9ycolor{OliveGreen}{+15+5}&=10y+20-y &color{Cerulean}{Combine:same-side:like:terms.} 9y+20&=9y+20 &color{Cerulean}{Combine:opposite-side:like:terms.} 9y+20color{Cerulean}{-9y}&=9y+20color{Cerulean}{-9y} 20&=20 quadcolor{Cerulean}{checkmark} &color{Cerulean}{True} end{aligned})

Answer:

(R). Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.

If it is hard to believe that any real number is a solution to the equation in the previous example, then choose your favorite real number, and substitute it in the equation to see that it leads to a true statement. Choose (x=7) and check:

Exercise (PageIndex{2})

Solve:

(−2(3x+1)−(x−3)=−7x+1).

Answer

(R)

Clearing Decimals and Fractions

The coefficients of linear equations may be any real number, even decimals and fractions. When decimals and fractions are used, it is possible to use the multiplication property of equality to clear the coefficients in a single step. If given decimal coefficients, then multiply by an appropriate power of 10 to clear the decimals. If given fractional coefficients, then multiply both sides of the equation by the least common multiple of the denominators (LCD).

Example (PageIndex{7})

Solve:

(2.3x+2.8=−1.2x+9.8).

Solution:

Notice that all decimal coefficients are expressed with digits in the tenths place; this suggests that we can clear the decimals by multiplying both sides by (10). Take care to distribute (10) to each term on both sides of the equation.

(egin{aligned} color{Cerulean}{10cdot }color{black}{(2.3x+2.8)} &=color{Cerulean}{10cdot}color{black}{-1.2x+9.8} &color{Cerulean}{Multiply:both:sides:by:10.} color{Cerulean}{10cdot }color{black}{2.3x+}color{Cerulean}{10cdot }color{black}{2.8}&=color{Cerulean}{10cdot }color{black}{(-1.2x)+}color{Cerulean}{10cdot}color{black}{9.8} 23x+28&=-12x+98 &color{Cerulean}{Integer:coefficients} 23x+28color{Cerulean}{+12x}&=-12x+98color{Cerulean}{+12x} &color{Cerulean}{Solve.} 35x+28&=98 35x+28color{Cerulean}{-28}&=98color{Cerulean}{-28} 35x&=70 frac{35x}{color{Cerulean}{35}}&=frac{70}{color{Cerulean}{35}} x&=2 end{aligned})

Answer:

The solution is (2).

Example (PageIndex{8})

Solve:

(frac{1}{3}x+frac{1}{5}=frac{1}{5}x−1).

Solution:

Clear the fractions by multiplying both sides by the least common multiple of the given denominators. In this case, the LCM((3, 5)=15).

Answer:

The solution is (-9).

It is important to know that these techniques only work for equations. Do not try to clear fractions when simplifying expressions. As a reminder

(egin{array}{c|c} {underline{color{Cerulean}{Expression}}}&{underline{color{Cerulean}{Equation}}} {frac{1}{2}x+frac{5}{3}}&{frac{1}{2}x+frac{5}{3}=0} end{array})

Solve equations and simplify expressions. If you multiply an expression by (6), you will change the problem. However, if you multiply both sides of an equation by 6, you obtain an equivalent equation.

(egin{array}{c|c} {underline{color{red}{Incorrect}}}&{underline{color{Cerulean}{Correct}}}{frac{1}{2}x+frac{5}{3}}&{frac{1}{2}x+frac{5}{3}=0} { eqcolor{red}{6cdot}color{black}{left( frac{1}{2}x+frac{5}{3} ight)}}&{color{Cerulean}{6cdot}color{black}{left( frac{1}{2}x+frac{5}{3} ight) =}color{Cerulean}{6cdot }color{black}{0}} {=3x+10quadcolor{red}{x}}&{3x+10=10quadcolor{Cerulean}{checkmark}} end{array})

Literal Equations (Linear Formulas)

Algebra lets us solve whole classes of applications using literal equations, or formulas. Formulas often have more than one variable and describe, or model, a particular real-world problem. For example, the familiar formula (D=rt) describes the distance traveled in terms of the average rate and time; given any two of these quantities, we can determine the third. Using algebra, we can solve the equation for any one of the variables and derive two more formulas.

(egin{aligned} D&=rt frac{D}{color{Cerulean}{r}}&=frac{rt}{color{Cerulean}{r}}&color{Cerulean}{Divide:both:sides:by:r.} frac{D}{r}&=t end{aligned})

If we divide both sides by (r), we obtain the formula (t=Dr). Use this formula to find the time, given the distance and the rate

(egin{aligned} D&=rt frac{D}{color{Cerulean}{t}}&=frac{rt}{color{Cerulean}{t}}&color{Cerulean}{Divide:both:sides:by:t.} frac{D}{t}&=r end{aligned})

If we divide both sides by (t), we obtain the formula (r=Dt). Use this formula to find the rate, given the distance traveled and the time it takes to travel that distance. Using the techniques learned up to this point, we now have three equivalent formulas relating distance, average rate, and time:

(D=rtqquad t=frac{D}{r}qquad r=frac{D}{t})

When given a literal equation, it is often necessary to solve for one of the variables in terms of the others. Use the properties of equality to isolate the indicated variable.

Example (PageIndex{9})

Solve for (a):

(P=2a+b).

Solution:

The goal is to isolate the variable (a).

(egin{aligned} P&=2a+b Pcolor{Cerulean}{-b}&=2a+bcolor{Cerulean}{-b} &color{Cerulean}{Subtract:b:from:both:sides.} P-b&=2a frac{P-b}{color{Cerulean}{2}}&=frac{2a}{color{Cerulean}{2}} &color{Cerulean}{Divide:both:sides:by:2.} frac{P-b}{2}&=a end{aligned})

Answer:

(a=frac{P-b}{2})

Example (PageIndex{10})

Solve for (y):

(z=frac{x+y}{2}).

Solution:

The goal is to isolate the variable (y).

(egin{aligned} z&=frac{x+y}{2} color{Cerulean}{2cdot}color{black}{z}&=color{Cerulean}{2cdot}color{black}{frac{x+y}{2}}&color{Cerulean}{Multiply:both:sides:by:2.} 2z&=x+y 2zcolor{Cerulean}{-x}&=x+ycolor{Cerulean}{-x}&color{Cerulean}{Subtract:x:from:both:sides.} 2z-x&=y end{aligned})

Answer:

(y=2z-x)

Exercise (PageIndex{3})

Solve for (b):

(2a−3b=c).

Answer

(b=frac{2a−c}{3})

Key Takeaways

  • Solving general linear equations involves isolating the variable, with coefficient (1), on one side of the equal sign.
  • The steps for solving linear equations are:
    • Simplify both sides of the equation and combine all same-side like terms.
    • Combine opposite-side like terms to obtain the variable term on one side of the equal sign and the constant term on the other.
    • Divide or multiply as needed to isolate the variable.
    • Check the answer.
  • Most linear equations that you will encounter are conditional and have one solution.
  • If solving a linear equation leads to a true statement like (0 = 0), then the equation is an identity and the solution set consists of all real numbers, (R).
  • If solving a linear equation leads to a false statement like (0 = 5), then the equation is a contradiction and there is no solution, (∅).
  • Clear fractions by multiplying both sides of a linear equation by the least common multiple of all the denominators. Distribute and multiply all terms by the LCD to obtain an equivalent equation with integer coefficients.
  • Given a formula, solve for any variable using the same techniques for solving linear equations. This works because variables are simply representations of real numbers.

Exercise (PageIndex{4}) Checking for Solutions

Is the given value a solution to the linear equation?

  1. (2(3x+5)−6=3x−8; x=−4 )
  2. (−x+17−8x=9−x; x=−1 )
  3. (4(3x−7)−3(x+2)=−1; x=frac{1}{3})
  4. (−5−2(x−5)=−(x+3); x=−8 )
  5. (7−2(frac{1}{2}x−6)=x−1; x=10 )
  6. (3x−frac{2}{3}(9x−2)=0; x=frac{4}{9})
Answer

1. Yes

3. No

5. Yes

Exercise (PageIndex{5}) Solving Linear Equations

Solve.

  1. (4x−7=7x+5)
  2. (−5x+3=−8x−9)
  3. (3x−5=2x−17)
  4. (−2y−52=3y+13)
  5. (−4x+2=7x−20)
  6. (4x−3=6x−15)
  7. (9x−25=12x−25)
  8. (12y+15=−6y+23)
  9. (1.2x−0.7=3x+4.7)
  10. (2.1x+6.1=−1.3x+4.4)
  11. (2.02x+4.8=14.782−1.2x)
  12. (−3.6x+5.5+8.2x=6.5+4.6x)
  13. (frac{1}{2}x−frac{2}{3}=x+frac{1}{5})
  14. (frac{1}{3}x−frac{1}{2}=−frac{1}{4}x−frac{1}{3})
  15. (−frac{1}{10}y+frac{2}{5}=frac{1}{5}y+frac{3}{10})
  16. (x−frac{20}{3}=frac{5}{2}x+frac{5}{6})
  17. (frac{2}{3}y+frac{1}{2}=frac{5}{8}y+frac{37}{24})
  18. (frac{1}{3}+frac{4}{3}x=frac{10}{7}x+frac{1}{3}−frac{2}{21}x)
  19. (frac{8}{9}−frac{11}{18}x=frac{7}{6}−12x)
  20. (frac{1}{3}−9x=frac{4}{9}+frac{1}{2}x)
  21. (12x−5+9x=44)
  22. (10−6x−13=12)
  23. (−2+4x+9=7x+8−2x)
  24. (20x−5+12x=6−x+7)
  25. (3a+5−a=2a+7)
  26. (−7b+3=2−5b+1−2b)
  27. (7x−2+3x=4+2x−2)
  28. (−3x+8−4x+2=10)
  29. (6x+2−3x=−2x−13)
  30. (3x−0.75+0.21x=1.24x+7.13)
  31. (−x−2+4x=5+3x−7)
  32. (−2y−5=8y−6−10y)
  33. (frac{1}{10}x−frac{1}{3}=frac{1}{30}−frac{1}{15}x−frac{7}{15})
  34. (frac{5}{8}−frac{4}{3}x+frac{1}{3}=−frac{3}{9}x−frac{1}{4}+frac{1}{3}x)
Answer

1. (−4)

3. (−12)

5. (2)

7. (0)

9. (−3)

11. (3.1)

13. (−frac{26}{15})

15. (frac{1}{3})

17. (25)

19. (−frac{5}{2})

21. (frac{7}{3})

23. (−1)

25. (∅)

27. (frac{1}{2})

29. (−3)

31. (R)

33. (−frac{3}{5})

Exercise (PageIndex{6}) Solving Linear Equations Involving Parentheses

Solve.

  1. (−5(2y−3)+2=12)
  2. (3(5x+4)+5x=−8)
  3. (4−2(x−5)=−2)
  4. (10−5(3x+1)=5(x−4))
  5. (9−(x+7)=2(x−1))
  6. (−5(2x−1)+3=−12)
  7. (3x−2(x+1)=x+5)
  8. (5x−3(2x−1)=2(x−3))
  9. (−6(x−1)−3x=3(x+8))
  10. (−frac{3}{5}(5x+10)=frac{1}{2}(4x−12))
  11. (3.1(2x−3)+0.5=22.2)
  12. (4.22−3.13(x−1)=5.2(2x+1)−11.38)
  13. (6(x−2)−(7x−12)=14)
  14. (−9(x−3)−3x=−3(4x+9))
  15. (3−2(x+4)=−3(4x−5))
  16. (12−2(2x+1)=4(x−1))
  17. (3(x+5)−2(2x+3)=7x+9)
  18. (3(2x−1)−4(3x−2)=−5x+10)
  19. (−3(2a−3)+2=3(a+7))
  20. (−2(5x−3)−1=5(−2x+1))
  21. (frac{1}{2}(2x+1)−frac{1}{4}(8x+2)=3(x−4))
  22. (−frac{2}{3}(6x−3)−frac{1}{2}=frac{3}{2}(4x+1))
  23. (frac{1}{2}(3x−1)+frac{1}{3}(2x−5)=0)
  24. (frac{1}{3}(x−2)+frac{1}{5}=frac{1}{9}(3x+3))
  25. (−2(2x−7)−(x+3)=6(x−1))
  26. (10(3x+5)−5(4x+2)=2(5x+20))
  27. (2(x−3)−6(2x+1)=−5(2x−4))
  28. (5(x−2)−(4x−1)=−2(3−x))
  29. (6(3x−2)−(12x−1)+4=0)
  30. (−3(4x−2)−(9x+3)−6x=0)
Answer

1. (frac{1}{2})

3. (8)

5. (frac{4}{3})

7. (∅)

9. (−frac{3}{2})

11. (5)

13. (−14)

15. (2)

17. (0)

19. (−frac{10}{9})

21. (3)

23. (1)

25. (frac{17}{11})

27. (∅)

29. (frac{7}{6})

Exercise (PageIndex{7}) Literal Equations

Solve for the indicated variable.

  1. Solve for (w): (A=l⋅w).
  2. Solve for (a): (F=ma).
  3. Solve for (w): (P=2l+2w).
  4. Solve for (r): (C=2πr).
  5. Solve for (b): (P=a+b+c).
  6. Solve for (C): (F=frac{9}{5}C+32).
  7. Solve for (h): (A=frac{1}{2}bh).
  8. Solve for (t): (I=Prt).
  9. Solve for (y): (ax+by=c).
  10. Solve for (h): (S=2πr^{2}+2πrh).
  11. Solve for (x): (z=frac{2x+y}{5}).
  12. Solve for (c): (a=3b−frac{2c}{3}).
  13. Solve for (b): (y=mx+b).
  14. Solve for (m): (y=mx+b).
  15. Solve for (y): (3x−2y=6).
  16. Solve for (y): (−5x+2y=12).
  17. Solve for (y): (frac{x}{3}−frac{y}{5}=1).
  18. Solve for (y): (frac{3}{4}x−frac{1}{5}y=frac{1}{2}).
Answer

1. (w=frac{A}{l})

3. (w=frac{P−2l}{2})

5. (b=P−a−c )

7. (h=frac{2A}{b})

9. (y=frac{−ax+c}{b})

11. (x=frac{5z−y}{2})

13. (b=y−mx)

15. (y=frac{3x−6}{2})

17. (y=frac{5x−15}{3})

Exercise (PageIndex{8}) Literal Equations

Translate the following sentences into linear equations and then solve.

  1. The sum of (3x) and (5) is equal to the sum of (2x) and (7).
  2. The sum of (−5x) and (6) is equal to the difference of (4x) and (2).
  3. The difference of (5x) and (25) is equal to the difference of (3x) and (51).
  4. The sum of (frac{1}{2}x) and (frac{3}{4}) is equal to (frac{2}{3}x).
  5. A number (n) divided by (5) is equal to the sum of twice the number and (3).
  6. Negative ten times a number (n) is equal to the sum of three times the number and (13).
Answer

1. (3x+5=2x+7); (x=2)

3. (5x−25=3x−51); (x=−13)

5. (frac{n}{5}=2n+3); (n=−frac{5}{3})

Exercise (PageIndex{9}) Discussion Board Topics

  1. What is the origin of the word algebra?
  2. What is regarded as the main business of algebra?
  3. Why is solving equations such an important algebra topic?
  4. Post some real-world linear formulas not presented in this section.
  5. Research and discuss the contributions of Diophantus of Alexandria.
  6. Create an identity or contradiction of your own and share on the discussion board. Provide a solution and explain how you found it.
Answer

1. Answers may vary

3. Answers may vary

5. Answers may vary


2.4: Solving Linear Equations- Part II

We’ll start off the solving portion of this chapter by solving linear equations. A linear equation is any equation that can be written in the form

where (a) and (b) are real numbers and (x) is a variable. This form is sometimes called the standard form of a linear equation. Note that most linear equations will not start off in this form. Also, the variable may or may not be an (x) so don’t get too locked into always seeing an (x) there.

To solve linear equations we will make heavy use of the following facts.

    If (a = b) then (a + c = b + c) for any (c). All this is saying is that we can add a number, (c), to both sides of the equation and not change the equation.

These facts form the basis of almost all the solving techniques that we’ll be looking at in this chapter so it’s very important that you know them and don’t forget about them. One way to think of these rules is the following. What we do to one side of an equation we have to do to the other side of the equation. If you remember that then you will always get these facts correct.

In this section we will be solving linear equations and there is a nice simple process for solving linear equations. Let’s first summarize the process and then we will work some examples.

Process for Solving Linear Equations

  1. If the equation contains any fractions use the least common denominator to clear the fractions. We will do this by multiplying both sides of the equation by the LCD.

Also, if there are variables in the denominators of the fractions identify values of the variable which will give division by zero as we will need to avoid these values in our solution.

Note that we usually just divide both sides of the equation by the coefficient if it is an integer or multiply both sides of the equation by the reciprocal of the coefficient if it is a fraction.

Let’s take a look at some examples.

  1. (3left( ight) = 2left( < - 6 - x> ight) - 2x)
  2. (displaystyle frac<><3>+ 1 = frac<<2m>><7>)
  3. (displaystyle frac<5><<2y - 6>> = frac<<10 - y>><<- 6y + 9>>)
  4. (displaystyle frac<<2z>><> = frac<3><> + 2)

For this problem there are no fractions so we don’t need to worry about the first step in the process. The next step tells to simplify both sides. So, we will clear out any parenthesis by multiplying the numbers through and then combine like terms.

[egin3left( ight) & = 2left( < - 6 - x> ight) - 2x 3x + 15 & = - 12 - 2x - 2x 3x + 15 & = - 12 - 4xend]

The next step is to get all the (x)’s on one side and all the numbers on the other side. Which side the (x)’s go on is up to you and will probably vary with the problem. As a rule of thumb, we will usually put the variables on the side that will give a positive coefficient. This is done simply because it is often easy to lose track of the minus sign on the coefficient and so if we make sure it is positive we won’t need to worry about it.

So, for our case this will mean adding 4(x) to both sides and subtracting 15 from both sides. Note as well that while we will actually put those operations in this time we normally do these operations in our head.

The next step says to get a coefficient of 1 in front of the (x). In this case we can do this by dividing both sides by a 7.

Now, if we’ve done all of our work correct (x = - frac<<27>><7>) is the solution to the equation.

The last and final step is to then check the solution. As pointed out in the process outline we need to check the solution in the original equation. This is important, because we may have made a mistake in the very first step and if we did and then checked the answer in the results from that step it may seem to indicate that the solution is correct when the reality will be that we don’t have the correct answer because of the mistake that we originally made.

The problem of course is that, with this solution, that checking might be a little messy. Let’s do it anyway.

So, we did our work correctly and the solution to the equation is,

Note that we didn’t use the solution set notation here. For single solutions we will rarely do that in this class. However, if we had wanted to the solution set notation for this problem would be,

Before proceeding to the next problem let’s first make a quick comment about the “messiness’ of this answer. Do NOT expect all answers to be nice simple integers. While we do try to keep most answer simple often they won’t be so do NOT get so locked into the idea that an answer must be a simple integer that you immediately assume that you’ve made a mistake because of the “messiness” of the answer.

Okay, with this one we won’t be putting quite as much explanation into the problem.

In this case we have fractions so to make our life easier we will multiply both sides by the LCD, which is 21 in this case. After doing that the problem will be very similar to the previous problem. Note as well that the denominators are only numbers and so we won’t need to worry about division by zero issues.

Let’s first multiply both sides by the LCD.

[egin21left( > <3>+ 1> ight) & = left( ><7>> ight)21 21left( ><3>> ight) + 21left( 1 ight) & = left( ><7>> ight)21 7left( ight) + 21 & = left( <2m> ight)left( 3 ight)end]

Be careful to correctly distribute the 21 through the parenthesis on the left side. Everything inside the parenthesis needs to be multiplied by the 21 before we simplify. At this point we’ve got a problem that is similar the previous problem and we won’t bother with all the explanation this time.

[egin7left( ight) + 21 & = left( <2m> ight)left( 3 ight) 7m - 14 + 21 & = 6m 7m + 7 & = 6m m & = - 7end]

So, it looks like (m = - 7) is the solution. Let’s verify it to make sure.

[eginfrac<< - 7 - 2>> <3>+ 1 & mathop = limits^? frac <<2left( < - 7> ight)>><7> frac<< - 9>> <3>+ 1 & mathop = limits^? - frac<<14>><7> - 3 + 1 &mathop = limits^? - 2 - 2 & = - 2hspace<0.5in>>end]

This one is similar to the previous one except now we’ve got variables in the denominator. So, to get the LCD we’ll first need to completely factor the denominators of each rational expression.

So, it looks like the LCD is (2 ight)^2>). Also note that we will need to avoid (y = 3) since if we plugged that into the equation we would get division by zero.

Now, outside of the (y)’s in the denominator this problem works identical to the previous one so let’s do the work.

Now the solution is not (y = 3) so we won’t get division by zero with the solution which is a good thing. Finally, let’s do a quick verification.

In this case it looks like the LCD is (left( ight)left( ight)) and it also looks like we will need to avoid (z = - 3) and (z = 10) to make sure that we don’t get division by zero.

Let’s get started on the work for this problem.

[eginleft( ight)left( ight)left( ><>> ight) & = left( <> + 2> ight)left( ight)left( ight) 2zleft( ight) & = 3left( ight) + 2left( ight)left( ight) 2 - 20z & = 3z + 9 + 2left( <- 7z - 30> ight)end]

At this point let’s pause and acknowledge that we’ve got a z 2 in the work here. Do not get excited about that. Sometimes these will show up temporarily in these problems. You should only worry about it if it is still there after we finish the simplification work.

So, let’s finish the problem.

Notice that the z 2 did in fact cancel out. Now, if we did our work correctly (z = frac<<17>><3>) should be the solution since it is not either of the two values that will give division by zero. Let’s verify this.

The checking can be a little messy at times, but it does mean that we KNOW the solution is correct.

Okay, in the last couple of parts of the previous example we kept going on about watching out for division by zero problems and yet we never did get a solution where that was an issue. So, we should now do a couple of those problems to see how they work.

The first step is to factor the denominators to get the LCD.

So, the LCD is (left( ight)left( ight)) and we will need to avoid (x = - 2) and (x = - 3) so we don’t get division by zero.

Here is the work for this problem.

[eginleft( ight)left( ight)left( <>> ight) & = left( > < ight)left( ight)>>> ight)left( ight)left( ight) 2left( ight) & = - x 2x + 6 & = - x 3x & = - 6 x & = - 2end]

So, we get a “solution” that is in the list of numbers that we need to avoid so we don’t get division by zero and so we can’t use it as a solution. However, this is also the only possible solution. That is okay. This just means that this equation has no solution.

The LCD for this equation is (x + 1) and we will need to avoid (x = - 1) so we don’t get division by zero. Here is the work for this equation.

[eginleft( <>> ight)left( ight) & = left( <4 - frac<<2x>><>> ight)left( ight) 2 & = 4left( ight) - 2x 2 & = 4x + 4 - 2x 2 & = 2x + 4 - 2 & = 2x - 1 & = xend]

So, we once again arrive at the single value of (x) that we needed to avoid so we didn’t get division by zero. Therefore, this equation has no solution.

So, as we’ve seen we do need to be careful with division by zero issues when we start off with equations that contain rational expressions.

At this point we should probably also acknowledge that provided we don’t have any division by zero issues (such as those in the last set of examples) linear equations will have exactly one solution. We will never get more than one solution and the only time that we won’t get any solutions is if we run across a division by zero problems with the “solution”.

Before leaving this section we should note that many of the techniques for solving linear equations will show up time and again as we cover different kinds of equations so it very important that you understand this process.


2.4: Solving Linear Equations- Part II

The first special case of first order differential equations that we will look at is the linear first order differential equation. In this case, unlike most of the first order cases that we will look at, we can actually derive a formula for the general solution. The general solution is derived below. However, we would suggest that you do not memorize the formula itself. Instead of memorizing the formula you should memorize and understand the process that I'm going to use to derive the formula. Most problems are actually easier to work by using the process instead of using the formula.

So, let's see how to solve a linear first order differential equation. Remember as we go through this process that the goal is to arrive at a solution that is in the form (y = yleft( t ight)). It's sometimes easy to lose sight of the goal as we go through this process for the first time.

In order to solve a linear first order differential equation we MUST start with the differential equation in the form shown below. If the differential equation is not in this form then the process we’re going to use will not work.

Where both (p(t)) and (g(t)) are continuous functions. Recall that a quick and dirty definition of a continuous function is that a function will be continuous provided you can draw the graph from left to right without ever picking up your pencil/pen. In other words, a function is continuous if there are no holes or breaks in it.

Now, we are going to assume that there is some magical function somewhere out there in the world, (mu left( t ight)), called an integrating factor. Do not, at this point, worry about what this function is or where it came from. We will figure out what (mu left( t ight)) is once we have the formula for the general solution in hand.

So, now that we have assumed the existence of (mu left( t ight)) multiply everything in (eqref) by (mu left( t ight)). This will give.

[eginmu left( t ight)frac<><

> + mu left( t ight)pleft( t ight)y = mu left( t ight)gleft( t ight) label end]

Now, this is where the magic of (mu left( t ight)) comes into play. We are going to assume that whatever (mu left( t ight)) is, it will satisfy the following.

[eginmu left( t ight)pleft( t ight) = mu 'left( t ight) label end]

Again do not worry about how we can find a (mu left( t ight)) that will satisfy (eqref). As we will see, provided (p(t)) is continuous we can find it. So substituting (eqref) we now arrive at.

[eginmu left( t ight)frac<><

> + mu 'left( t ight)y = mu left( t ight)gleft( t ight) label end]

At this point we need to recognize that the left side of (eqref) is nothing more than the following product rule.

So we can replace the left side of (eqref) with this product rule. Upon doing this (eqref) becomes

Now, recall that we are after (y(t)). We can now do something about that. All we need to do is integrate both sides then use a little algebra and we'll have the solution. So, integrate both sides of (eqref) to get.

Note the constant of integration, (c), from the left side integration is included here. It is vitally important that this be included. If it is left out you will get the wrong answer every time.

The final step is then some algebra to solve for the solution, (y(t)).

Now, from a notational standpoint we know that the constant of integration, (c), is an unknown constant and so to make our life easier we will absorb the minus sign in front of it into the constant and use a plus instead. This will NOT affect the final answer for the solution. So with this change we have.

Again, changing the sign on the constant will not affect our answer. If you choose to keep the minus sign you will get the same value of (c) as we do except it will have the opposite sign. Upon plugging in (c) we will get exactly the same answer.

There is a lot of playing fast and loose with constants of integration in this section, so you will need to get used to it. When we do this we will always to try to make it very clear what is going on and try to justify why we did what we did.

So, now that we’ve got a general solution to (eqref) we need to go back and determine just what this magical function (mu left( t ight)) is. This is actually an easier process than you might think. We’ll start with (eqref).

[mu left( t ight)pleft( t ight) = mu 'left( t ight)]

Divide both sides by (mu left( t ight)),

Now, hopefully you will recognize the left side of this from your Calculus I class as nothing more than the following derivative.

As with the process above all we need to do is integrate both sides to get.

You will notice that the constant of integration from the left side, (k), had been moved to the right side and had the minus sign absorbed into it again as we did earlier. Also note that we’re using (k) here because we’ve already used (c) and in a little bit we’ll have both of them in the same equation. So, to avoid confusion we used different letters to represent the fact that they will, in all probability, have different values.

Exponentiate both sides to get (mu left( t ight)) out of the natural logarithm.

Now, it’s time to play fast and loose with constants again. It is inconvenient to have the (k) in the exponent so we’re going to get it out of the exponent in the following way.

Now, let’s make use of the fact that (k) is an unknown constant. If (k) is an unknown constant then so is (<<f>^k>) so we might as well just rename it (k) and make our life easier. This will give us the following.

So, we now have a formula for the general solution, (eqref), and a formula for the integrating factor, (eqref). We do have a problem however. We’ve got two unknown constants and the more unknown constants we have the more trouble we’ll have later on. Therefore, it would be nice if we could find a way to eliminate one of them (we’ll not be able to eliminate both….).

This is actually quite easy to do. First, substitute (eqref) into (eqref) and rearrange the constants.

So, (eqref) can be written in such a way that the only place the two unknown constants show up is a ratio of the two. Then since both (c) and (k) are unknown constants so is the ratio of the two constants. Therefore we’ll just call the ratio (c) and then drop (k) out of (eqref) since it will just get absorbed into (c) eventually.

The solution to a linear first order differential equation is then

Now, the reality is that (eqref) is not as useful as it may seem. It is often easier to just run through the process that got us to (eqref) rather than using the formula. We will not use this formula in any of our examples. We will need to use (eqref) regularly, as that formula is easier to use than the process to derive it.

Solution Process

The solution process for a first order linear differential equation is as follows.

    Put the differential equation in the correct initial form, (eqref).

Let’s work a couple of examples. Let’s start by solving the differential equation that we derived back in the Direction Field section.

First, we need to get the differential equation in the correct form.

From this we can see that (p(t)=0.196) and so (mu left( t ight)) is then.

Note that officially there should be a constant of integration in the exponent from the integration. However, we can drop that for exactly the same reason that we dropped the (k) from (eqref).

Now multiply all the terms in the differential equation by the integrating factor and do some simplification.

Integrate both sides and don't forget the constants of integration that will arise from both integrals.

Okay. It’s time to play with constants again. We can subtract (k) from both sides to get.

Both (c) and (k) are unknown constants and so the difference is also an unknown constant. We will therefore write the difference as (c). So, we now have

From this point on we will only put one constant of integration down when we integrate both sides knowing that if we had written down one for each integral, as we should, the two would just end up getting absorbed into each other.

The final step in the solution process is then to divide both sides by (<<f>^<0.196t>>) or to multiply both sides by (<<f>^< - 0.196t>>). Either will work, but we usually prefer the multiplication route. Doing this gives the general solution to the differential equation.

From the solution to this example we can now see why the constant of integration is so important in this process. Without it, in this case, we would get a single, constant solution, (v(t)=50). With the constant of integration we get infinitely many solutions, one for each value of (c).

Back in the direction field section where we first derived the differential equation used in the last example we used the direction field to help us sketch some solutions. Let's see if we got them correct. To sketch some solutions all we need to do is to pick different values of (c) to get a solution. Several of these are shown in the graph below.

So, it looks like we did pretty good sketching the graphs back in the direction field section.

Now, recall from the Definitions section that the Initial Condition(s) will allow us to zero in on a particular solution. Solutions to first order differential equations (not just linear as we will see) will have a single unknown constant in them and so we will need exactly one initial condition to find the value of that constant and hence find the solution that we were after. The initial condition for first order differential equations will be of the form

Recall as well that a differential equation along with a sufficient number of initial conditions is called an Initial Value Problem (IVP).

To find the solution to an IVP we must first find the general solution to the differential equation and then use the initial condition to identify the exact solution that we are after. So, since this is the same differential equation as we looked at in Example 1, we already have its general solution.

Now, to find the solution we are after we need to identify the value of (c) that will give us the solution we are after. To do this we simply plug in the initial condition which will give us an equation we can solve for (c). So, let's do this

[48 = vleft( 0 ight) = 50 + chspace <0.25in>Rightarrow hspace<0.25in>c = - 2]

So, the actual solution to the IVP is.

A graph of this solution can be seen in the figure above.

Let’s do a couple of examples that are a little more involved.

Rewrite the differential equation to get the coefficient of the derivative a one.

Now find the integrating factor.

Can you do the integral? If not rewrite tangent back into sines and cosines and then use a simple substitution. Note that we could drop the absolute value bars on the secant because of the limits on (x). In fact, this is the reason for the limits on (x). Note as well that there are two forms of the answer to this integral. They are equivalent as shown below. Which you use is really a matter of preference.

Also note that we made use of the following fact.

This is an important fact that you should always remember for these problems. We will want to simplify the integrating factor as much as possible in all cases and this fact will help with that simplification.

Now back to the example. Multiply the integrating factor through the differential equation and verify the left side is a product rule. Note as well that we multiply the integrating factor through the rewritten differential equation and NOT the original differential equation. Make sure that you do this. If you multiply the integrating factor through the original differential equation you will get the wrong solution!

[eginsec left( x ight)y' + sec left( x ight) an left( x ight)y & = 2sec left( x ight)left( x ight)sin left( x ight) - left( x ight) ight)^prime > & = 2cos left( x ight)sin left( x ight) - left( x ight)end]

Note the use of the trig formula (sin left( <2 heta > ight) = 2sin heta cos heta ) that made the integral easier. Next, solve for the solution.

Finally, apply the initial condition to find the value of (c).

[yleft( x ight) = - frac<1><2>cos left( x ight)cos left( <2x> ight) - sin left( x ight) + 7cos left( x ight)]

Below is a plot of the solution.

[t,y' + 2y = - t + 1hspace<0.25in>yleft( 1 ight) = frac<1><2>]

First, divide through by the t to get the differential equation into the correct form.

Now let’s get the integrating factor, (mu left( t ight)).

Now, we need to simplify (mu left( t ight)). However, we can’t use (eqref) yet as that requires a coefficient of one in front of the logarithm. So, recall that

and rewrite the integrating factor in a form that will allow us to simplify it.

We were able to drop the absolute value bars here because we were squaring the (t), but often they can’t be dropped so be careful with them and don’t drop them unless you know that you can. Often the absolute value bars must remain.

Now, multiply the rewritten differential equation (remember we can’t use the original differential equation here…) by the integrating factor.

Integrate both sides and solve for the solution.

Finally, apply the initial condition to get the value of (c).

[frac<1> <2>= yleft( 1 ight) = frac<1> <4>- frac<1> <3>+ frac<1> <2>+ chspace <0.25in>Rightarrow hspace<0.25in>,,,c = frac<1><<12>>]

Here is a plot of the solution.

First, divide through by (t) to get the differential equation in the correct form.

Now that we have done this we can find the integrating factor, (mu left( t ight)).

Do not forget that the “-” is part of (p(t)). Forgetting this minus sign can take a problem that is very easy to do and turn it into a very difficult, if not impossible problem so be careful!

Now, we just need to simplify this as we did in the previous example.

Again, we can drop the absolute value bars since we are squaring the term.

Now multiply the differential equation by the integrating factor (again, make sure it’s the rewritten one and not the original differential equation).

Integrate both sides and solve for the solution.

[egin>yleft( t ight) & = int<<sin left( <2t> ight),dt>> + int<< - 1 + 4t,dt>> >yleft( t ight) & = - frac<1><2>cos left( <2t> ight) + frac<1><2>tsin left( <2t> ight) + frac<1><4>cos left( <2t> ight) - t + 2 + c yleft( t ight) & = - frac<1><2>cos left( <2t> ight) + frac<1><2>sin left( <2t> ight) + frac<1><4>cos left( <2t> ight) - + 2 + cend]

Apply the initial condition to find the value of (c).

Below is a plot of the solution.

Let’s work one final example that looks more at interpreting a solution rather than finding a solution.

First, divide through by a 2 to get the differential equation in the correct form.

[y' - frac<1><2>y = 2sin left( <3t> ight)]

Multiply (mu left( t ight))through the differential equation and rewrite the left side as a product rule.

Integrate both sides (the right side requires integration by parts – you can do that right?) and solve for the solution.

Apply the initial condition to find the value of (c) and note that it will contain (y_<0>) as we don’t have a value for that.

Now that we have the solution, let’s look at the long term behavior (i.e. (t o infty )) of the solution. The first two terms of the solution will remain finite for all values of (t). It is the last term that will determine the behavior of the solution. The exponential will always go to infinity as (t o infty ), however depending on the sign of the coefficient (c) (yes we’ve already found it, but for ease of this discussion we’ll continue to call it (c)). The following table gives the long term behavior of the solution for all values of (c).

Range of (c) Behavior of solution as(t o infty )
(c) < 0 (yleft( t ight) o - infty )
(c) = 0 (yleft( t ight)) remains finite
(c) > 0 (yleft( t ight) o infty )

This behavior can also be seen in the following graph of several of the solutions.

Now, because we know how (c) relates to (y_<0>) we can relate the behavior of the solution to (y_<0>). The following table give the behavior of the solution in terms of (y_<0>) instead of (c).

Note that for ( = - frac<<24>><<37>>) the solution will remain finite. That will not always happen.

Investigating the long term behavior of solutions is sometimes more important than the solution itself. Suppose that the solution above gave the temperature in a bar of metal. In this case we would want the solution(s) that remains finite in the long term. With this investigation we would now have the value of the initial condition that will give us that solution and more importantly values of the initial condition that we would need to avoid so that we didn’t melt the bar.


1.2.2. Direction Fields

* Refer to Maple file “Direction Fields”

When solving ODEs, there are many methods in plotting them. In this section, we will learn how to use three plot commands in the DEtools package to plot the solutions to ODEs. The dfieldplot command draws out a direction/slope field of the given function. The generic syntax of the command is as follows:

> with(DEtools):
> dfieldplot(differential equation, independent variable, x range, y range)

In order to express a differential equation, for example a function of y in relation to x, you must enter “diff(y(x),x).” In some tutorials, this can be expressed as “D(y)(x),” but for simplicities sake, we will use the former expression.
Try out the following:

The input should generate the above direction field. Unfortunately, you must plot differential equations using dfieldplot explicitly. Therefore, it is best to solve a given differential equation to express it explicitly if it is given implicitly.


Solving Linear Equations that contain Fractions

Solving Linear Equations that have Variables on both sides

Solving Linear Equations that contain Parentheses

General Procedure for Solving a Linear Equation: pg. 152

1. If fractions are present, multiply both sides of the equation by the LCD of all the fractions.
2. Simplify each side of the equation.
3. Use addition and subtraction to get all terms with the variable on one side and all constant terms
on the other side of the equation. Then, simplify each side of the equation.
4. Divide both sides of the equation by the coefficient of the variable.
5. Check your solution in the original equation.

These problems have exactly one solution.

These problems have no solution

These problems have every value of the variable as a solution. They are called an identity.

Review: When you solve a linear equation, if the simplified equation is:

A false statement, such as 8 = 1, the equation has no solution.
A true statement, such as 12 = 12, the equation is an identity.
Any value of the variable is a solution.
A true statement, such as x = 4, there is only one solution.


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Elementary Differential Equations

Office : ACD 114A
Phone : (860) 405-9294
Office Hours : on demand
Open Door Policy: You are welcome to drop by to discuss any aspect of the course, anytime, on the days I am on campus-- Tuesday, Thursday, and Friday.

MATH 2410 covers material from chapters of the textbook. The topics include first and second order equations, systems, Laplace transforms.


Class Meeting Times/Place: Tuesday, Thursday 2:00 - 3:15 p.m. Classroom ACD 206.

Due to Corona Virus outbreak, all the classes after the spring break will take place online at the usual class time, TuTh 2:00-3:15 p.m.

The textbook for the course is “A First Course in Differential Equations with Modeling Applications” by Dennis G. Zill, 11th edition.

The homework for Math 2410 is assigned at the end of each class and is collected every Thursday. The total weight of the homework grades is 150 point of the total 500 course points.

Exam Schedule: Exam 1: Thursday, February 21 ,  2:00 - 3:15p.m., Room: ACD 206
Exam 2: Thursday, March 28 ,  2:00 - 3:15p.m., Room: ACD 206
Final Exam: Tuesday, May 1, 1:00 - 3:00 p.m., Room: ACD 206

Grading Policy: Homework: 150, Exam 1: 100, Exam 2: 100, Final Exam: 150.



Date Chapter Topic Homework
Week 1 Tues. 1/21 1.1 Definitions and Terminology Ch. 1.1

Thur. 1/23 1.2
1.3
Initial-Value problems
Differential Equations as mathematical models
Ch. 1.2
Ch. 1.3





Week 2 Tues. 1/28 2.1 Solution curves without a solution (Direction fields/Autonomous). Ch. 2.1

Thur. 1/30 2.2 Phase portraits. Separable equations Ch. 2.2





Week 3 Tues. 2/4 2.3 Linear equations Ch. 2.3

Thur. 2/6 2.4 Exact equations. Ch. 2.4





Week 4 Tues. 2/11 2.5 Solutions by substitutions. Ch. 2.5

Thur. 2/13 2.6 Euler's method. Ch. 2.6





Week 5 Tues. 2/18
Review.
Practice Exam 1
Practice Exam 1. Solutions


Thur. 2/20
Exam 1





Week 6 Tues. 2/25 3.1 Linear models. Ch. 3.1

Thur. 2/27 3.2, 3.3 Linear models. Modeling with Systems of 1st order Equations. Ch. 3.2, 3.3





Week 7 Tues. 3/3 8.1 Linear models. Ch. 8.1

Thur. 3/5 8.2 Homogeneous linear systems. Ch. 8.2





Week 8 Tues. 3/10 8.2 Homogeneous linear systems. Repeated eigenvalues. Ch. 8.2

Thur. 3/12 8.2 Homogeneous linear systems. Complex eigenvalues. Ch. 8.2





Week 9 Tues. 3/17
Spring recess

Thur. 3/19
Spring recess





Week 10 Tues. 3/24 8.3 Nonhomogeneous linear systems. Ch. 8.3

Thur. 3/26 8.3 Nonhomogeneous linear systems. 8.3





Week 11 Tues. 3/31
Review.
Practice Exam 2
Practice Exam 2. Solutions

Thur. 4/2
Exam 2





Week 12 Tues. 4/7 4.1 Exam review. Linear equations, part 1.

Thur. 4/9 4.1 Linear equations. part 2. Ch. 4.1





Week 13 Tues. 4/14 4.2, 4.3 Reduction order. Homogeneous linear equations with constant coefficients. Ch. 4.1

Thur. 4/16 4.4 Undetermined coefficients. Superposition approach. Ch. 4.1





Week 14 Tues. 4/21 7.1 The Laplace Transform Ch. 4.3

Thur. 4/23 7.2 Inverse Transforms and Transforms of Derivatives. Ch. 4.4





Week 15 Tues. 4/28 7.3 Operation properties 1. Ch. 7.1
Thur. 4/30 7.4 Operation properties 2. Ch. 7.2





Week 16 Tues. 5/5
Final Exam, 1:00-3:00 p.m.
Practice Final Exam
Practice Final Exam. Solutions

This page is maintained by Dmitriy Leykekhman
Last modified: 4/28/2020


Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

Linear Equations in one Variable Exercise 14A – Selina Concise Mathematics Class 8 ICSE Solutions

Solve the following equations:
Question 1.
20 = 6 + 2x
Solution:

Question 2.
15 + x = 5x + 3
Solution:

Question 3.
= -7
Solution:

Question 4.
3a – 4 = 2(4 – a)
Solution:

Question 5.
3(b – 4) = 2(4 – b)
Solution:

Solution:

Solution:

Question 8.
5(8x + 3) = 9(4x + 7)
Solution:

Question 9.
3(x + 1) = 12 + 4(x – 1)
Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Question 16.
6(6x – 5) – 5 (7x – 8) = 12 (4 – x) + 1
Solution:

Question 17.
(x – 5) (x + 3) = (x – 7) (x + 4)
Solution:

Question 18.
(x – 5) 2 – (x + 2) 2 = -2
Solution:

Question 19.
(x – 1) (x + 6) – (x – 2) (x – 3) = 3
Solution:

Solution:

Solution:

Solution:

Solution:

Question 24.
Solve:
Hence, find the value of ‘a’, if + 5x = 8.
Solution:

Question 25.
Solve:
Hence find the value of ‘p’ if 2p – 2x + 1 = 0
Solution:

Question 26.
Solve:
Solution:

Question 27.
Solve:
Solution:

Linear Equations in one Variable Exercise 14B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Fifteen less than 4 times a number is 9. Find the number.
Solution:

Question 2.
If Megha’s age is increased by three times her age, the result is 60 years. Find her age
Solution:

Question 3.
28 is 12 less than 4 times a number. Find the number.
Solution:

Question 4.
Five less than 3 times a number is -20. Find the number.
Solution:

Question 5.
Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she?
Solution:

Question 6.
A number decreased by 30 is the same as 14 decreased by 3 times the number Find the number.
Solution:

Question 7.
A’s salary is same as 4 times B’s salary. If together they earn Rs.3,750 a month, find the salary of each.
Solution:

Question 8.
Separate 178 into two parts so that the first part is 8 less than twice the second part.
Solution:

Question 9.
Six more than one-fourth of a number is two-fifth of the number. Find the number.
Solution:

Question 10.
The length of a rectangle is twice its width. If its perimeter is 54 cm find its length.
Solution:

Question 11.
A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the origi¬nal rectangle.
Solution:

Question 12.
The sum of three consecutive odd numbers is 57. Find the numbers.
Solution:

Question 13.
A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.
Solution:

Question 14.
A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?
Solution:

Question 15.
A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.
Solution:

Question 16.
The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.
Solution:

Question 17.
Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of the greater by 1. Find the numbers.
Solution:

Question 18.
Three consecutive whole numbers are such that if they are divided by 5, 3 and 4 respectively the sum of the quotients is 40. Find the numbers.
Solution:

Question 19.
If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.
Solution:

Question 20.
The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.
Solution:

Linear Equations in one Variable Exercise 14C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Solve:
(i)
(ii)
(iii) (x + 2)(x + 3) + (x – 3)(x – 2) – 2x(x + 1) = 0
(iv)
(v) 13(x – 4) – 3(x – 9) – 5(x + 4) = 0
(vi) x + 7 –
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
Solution:






Question 2.
After 12 years, I shall be 3 times as old as 1 was 4 years ago. Find my present age.
Solution:

Question 3.
A man sold an article for 7396 and gained 10% on it. Find the cost price of the article
Solution:

Question 4.
The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.
Solution:

Question 5.
The sum of two numbers is 405 and their ratio is 8 : 7. Find the numbers.
Solution:

Question 6.
The ages of A and B are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.
Solution:

Question 7.
Find the number whose double is 45 greater than its half.
Solution:

Question 8.
The difference between the squares of two consecutive numbers is 31. Find the numbers.
Solution:

Question 9.
Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Solution:

Question 10.
The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator, and denominator both, the fraction becomes . Find the original fraction.
Solution:


Meets NCTM Standards:

Description: [NOTE: This video is only a 12 minute portion of the full 3 hour lesson. Please visit my website to purchase complete version] This lesson consists of providing you with a Self-Tutorial on how to solve typical linear word problems (story problems or applied problems). The tutor shows you how to solve for a specific variable in formulas. He also discusses how to covert a repeating decimal into a fraction (which was skipped in Basic Math: Lesson 6 -"Fractions") and will teach you how to convert units of measurement.

Examples of word problems done include:

Finding a number based on certain criteria.
Word problems involving some geometry (triangle, rectangle, circle).
Age problems.
Mixture problems.
Money problems (story of my life!).
Rate-Time-Distance problems.
Percent Equations/problems.
Ratio and Proportion (concepts and solving problems, including similar triangles).
Problems dealing with Unit Price.


2.4: Solving Linear Equations- Part II

Welcome to my online math tutorials and notes. The intent of this site is to provide a complete set of free online (and downloadable) notes and/or tutorials for classes that I teach at Lamar University. I've tried to write the notes/tutorials in such a way that they should be accessible to anyone wanting to learn the subject regardless of whether you are in my classes or not. In other words, they do not assume you've got any prior knowledge other than the standard set of prerequisite material needed for that class. In other words, it is assumed that you know Algebra and Trig prior to reading the Calculus I notes, know Calculus I prior to reading the Calculus II notes, etc. The assumptions about your background that I've made are given with each description below.

I'd like to thank Shane F, Fred J., Mike K. and David A. for all the typos that they've found and sent my way! I've tried to proof read these pages and catch as many typos as I could, however it just isn't possible to catch all of them when you are also the person who wrote the material. Fred, Mike and David have caught quite a few typos that I'd missed and been nice enough to send them my way. Thanks again Fred, Mike and David!

If you are one of my current students and are here looking for homework assignments I've got a set of links that will get you to the right pages listed here.

At present I've gotten the notes/tutorials for my Algebra (Math 1314), Calculus I (Math 2413), Calculus II (Math 2414), Calculus III (Math 3435) and Differential Equations (Math 3301) class online. I've also got a couple of Review/Extras available as well. Among the reviews/extras that I've got are an Algebra/Trig review for my Calculus Students, a Complex Number primer, a set of Common Math Errors, and some tips on How to Study Math.

I've made most of the pages on this site available for download as well. These downloadable versions are in pdf format. Each subject on this site is available as a complete download and in the case of very large documents I've also split them up into smaller portions that mostly correspond to each of the individual topics. To get the downloadable version of any topic navigate to that topic and then under the Download menu you will be presented an option to download the topic.

Here is a complete listing of all the subjects that are currently available on this site as well as brief descriptions of each.

Algebra Cheat Sheets - This is as many common algebra facts, properties, formulas, and functions that I could think of. There is also a page of common algebra errors included. There are two versions of the cheat sheet available. One is full sized and is currently four pages. The other version is a reduced version that contains exactly the same information as the full version except it has just been shrunk down so two pages print of the front and two pages print on the back of a single piece of paper.

Trig Cheat Sheets - Here is a set of common trig facts, properties and formulas. A unit circle (completely filled out) is also included. There are two versions of the cheat sheet available. One is full sized and is currently four pages. The other version is a reduced version that contains exactly the same information as the full version except it has just been shrunk down so two pages print of the front and two pages print on the back of a single piece of paper.

Calculus Cheat Sheets - These are a series of Calculus Cheat Sheets that covers most of a standard Calculus I course and a few topics from a Calculus II course. There are four different cheat sheets here. One contains all the information, one has just Limits information, one has just Derivatives information and the final one has just Integrals information. Each cheat sheets comes in two versions. One that is full sized and another that has been reduced, with exactly the same information as the full sized version, that prints two pages on the front and/or back of each page of paper.

Common Derivatives and Integrals - Here is a set of common derivatives and integrals that are used somewhat regularly in a Calculus I or Calculus II class. Also included are reminders on several integration techniques. here are two versions of the cheat sheet available. One is full sized and is currently four pages. The other version is a reduced version that contains exactly the same information as the full version except it has just been shrunk down so two pages print of the front and two pages print on the back of a single piece of paper.

Table of Laplace Transforms - Here is a list of Laplace transforms for a differential equations class. This table gives many of the commonly used Laplace transforms and formulas. It is currently two pages long with the first page being the Laplace transforms and the second being some information/facts about some of the entries.

All of the classes, with the exception of Differential Equations, have practice problems (with solutions) you can use for practice as well as a set of assignment problems (without solutions/answers) for instructors to use if they wish.

  • Preliminaries - Exponent Properties, Rational Exponents, Negative Exponents, Radicals, Polynomials, Factoring, Rational Expressions, Complex Numbers
  • Solving Equations and Inequalities - Linear Equations, Quadratic Equations, Completing the Square, Quadratic Formula, Applications of Linear and Quadratic Equations, Reducible to Quadratic Form, Equations with Radicals, Linear Inequalities, Polynomial & Rational Inequalities, Absolute Value Equations & Inequalities.
  • Graphing and Functions - Graphing Lines, Circles, and Piecewise Functions, Function Definition, Function Notation, Function Composition, Inverse Functions.
  • Common Graphs - Parabolas, Ellipses, Hyperbolas, Absolute Value, Square Root, Constant Function, Rational Functions, Shifts, Reflections, Symmetry.
  • Polynomial Functions - Dividing Polynomials, Zeroes/Roots of Polynomials, Finding Zeroes of Polynomials, Graphing Polynomials, Partial Fractions.
  • Exponential and Logarithm Functions - Exponential Functions, Logarithm Functions, Solving Exponential Functions, Solving Logarithm Functions, Applications.
  • Systems of Equations - Substitution Method, Elimination Method, Augmented Matrix, Nonlinear Systems.

The Algebra notes/tutorial assume that you've had some exposure to the basics of Algebra. In particular it is assumed that the exponents and factoring sections will be more of a review for you. Also, it is assumed that you've seen the basics of graphing equations. Graphing particular types of equations is covered extensively in the notes, however, it is assumed that you understand the basic coordinate system and how to plot points.

  • Algebra/Trig Review - Trig Functions and Equations, Exponential Functions and Equations, Logarithm Functions and Equations.
  • Limits - Concepts, Definition, Computing, One-Sided Limits, Continuity, Limits Involving Infinity, L'Hospitals Rule
  • Derivatives - Definition, Interpretations, Derivative Formulas, Power Rule, Product Rule, Quotient Rule, Chain Rule, Higher Order Derivatives, Implicit Differentiation, Logarithmic Differentiation, Derivatives of Trig Functions, Exponential Functions, Logarithm Functions, Inverse Trig Functions, and Hyperbolic Trig Functions.
  • Applications of Derivatives - Related Rates, Critical Points, Minimum and Maximum Values, Increasing/Decreasing Functions, Inflection Points, Concavity, Optimization
  • Integration - Definition, Indefinite Integrals, Definite Integrals, Substitution Rule, Evaluating Definite Integrals, Fundamental Theorem of Calculus
  • Applications of Integrals - Average Function Value, Area Between Curves, Solids of Revolution, Work.

The Calculus I notes/tutorial assume that you've got a working knowledge of Algebra and Trig. There is some review of a couple of Algebra and Trig topics, but for the most part it is assumed that you do have a decent background in Algebra and Trig. These notes assume no prior knowledge of Calculus.

  • Integration Techniques - Integration by Parts, Integrals Involving Trig Functions, Trig Substitutions, Integration using Partial Fractions, Integrals Involving Roots, Integrals Involving Quadratics, Integration Strategy, Improper Integrals, Comparison Test for Improper Integrals, and Approximating Definite Integrals.
  • Applications of Integrals - Arc Length, Surface Area, Center of Mass/Centroid, Hydrostatic Pressure and Force, Probability.
  • Parametric Equations and Polar Coordinates - Parametric Equations & Curves, Calculus with Parametric Equations (Tangents, Areas, Arc Length and Surface Area), Polar Coordinates, Calculus with Polar Coordinates (Tangents, Areas, Arc Length and Surface Area).
  • Sequences and Series - Sequences, Series, Convergence/Divergence of Series, Absolute Series, Integral Test, Comparison Test, Limit Comparison Test, Alternating Series Test, Ratio Test, Root Test, Estimating the Value of a Series, Power Series, Taylor Series, Binomial Series
  • Vectors - Basics, Magnitude, Unit Vector, Arithmetic, Dot Product, Cross Product, Projection
  • Three Dimensional Coordinate System - Equations of Lines, Equations of Planes, Quadratic Surfaces, Functions of Multiple Variables, Vector Functions, Limits, Derivatives, and Integrals of Vector Functions, Tangent Vectors, Normal Vectors, Binormal Vectors, Curvature, Cylindrical Coordinates, Spherical Coordinates

The Calculus II notes/tutorial assume that you've got a working knowledge Calculus I, including Limits, Derivatives, and Integration (up to basic substitution). It is also assumed that you have a fairly good knowledge of Trig. Several topics rely heavily on trig and knowledge of trig functions.

  • Three Dimensional Coordinate System - Equations of Lines, Equations of Planes, Quadratic Surfaces, Functions of Multiple Variables, Vector Functions, Limits, Derivatives, and Integrals of Vector Functions, Tangent Vectors, Normal Vectors, Binormal Vectors, Curvature, Cylindrical Coordinates, Spherical Coordinates
  • Partial Derivatives - Limits, Partial Derivatives, Higher Order Partial Derivatives, Differentials, Chain Rule, Directional Derivatives, Gradient.
  • Applications of Partial Derivatives - Tangent Plane, Normal Line, Relative Extrema, Absolute Extrema, Optimization, Lagrange Multipliers.
  • Multiple Integrals - Iterated Integrals, Double Integrals, Double Integrals in Polar Coordinates, Triple Integrals, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Change of Variables, Surface Area.
  • Line Integrals - Vector Fields, Line Integrals With Respect to Arc Length, Line Integrals With Respect to x and y, Line Integrals of Vector Fields, Fundamental Theorem of Line Integrals, Conservative Vector Fields, Potential Functions, Green's Theorem, Curl, Divergence.
  • Surface Integrals - Parametric Surfaces, Surface Integrals, Surface Integrals of Vector Fields, Stokes' Theorem, Divergence Theorem.

The Calculus III notes/tutorial assume that you've got a working knowledge Calculus I, including limits, derivatives and integration. It also assumes that the reader has a good knowledge of several Calculus II topics including some integration techniques, parametric equations, vectors, and knowledge of three dimensional space.

  • First Order Differential Equations - Linear Equations, Separable Equations, Exact Equations, Equilibrium Solutions, Modeling Problems.
  • Second Order Differential Equations - Homogeneous and Nonhomogeneous Second Order Differential Equations, Fundamental Set of Solutions, Undetermined Coefficients, Variation of Parameters, Mechanical Vibrations
  • Laplace Transforms - Definition, Inverse Transforms, Step Functions, Heaviside Functions, Dirac-Delta Function, Solving IVP's, Nonhomogeneous IVP, Nonconstant Coefficient IVP, Convolution Integral.
  • Systems of Differential Equations - Matrix Form, Eigenvalues/Eigenvectors, Phase Plane, Nonhomogeneous Systems, Laplace Transforms.
  • Series Solutions - Series Solutions, Euler Differential Equations.
  • Higher Order Differential Equations - n th order differential equations, Undetermined Coefficients, Variation of Parameters, 3 x 3 Systems of Differential Equations.
  • Boundary Value Problems & Fourier Series - Boundary Value Problems, Eigenvalues and Eigenfunctions, Orthogonal Functions, Fourier Sine Series, Fourier Cosine Series, Fourier Series.
  • Partial Differential Equations - Heat Equation, Wave Equation, Laplace's Equation, Separation of Variables.

These notes assume no prior knowledge of differential equations. A good grasp of Calculus is required however. This includes a working knowledge of differentiation and integration.

Not all the topics covered in an Algebra or Trig class are covered in this review. I've mostly covered topics that are of particular importance to students in a Calculus class. I have included a couple of topics that are not that important to a Calculus class, but students do seem to have trouble with on occasion. As time permits I will be adding more sections as well.

The review is in the form of a problem set with the first solution containing detailed information on how to work that type of problem. Later solutions are usually not as detailed, but may contain more/new information as required.

Note that this primer does assume that you've at least seen some complex numbers prior to reading. The purpose of this document is go a little beyond what most people see when the first are introduced to complex numbers in say a College Algebra class. Also, this document is in no way intended to be a complete picture of complex numbers nor do I cover all the concepts involved (that's a whole class in and of itself).

This portion of the site should be of interest to anyone looking for common math errors. If you aren't in a Calculus class or haven't taken Calculus you should just ignore the last section.

How To Study Math - This is a short section with some advice on how to best study mathematics.


Watch the video: Solve the linear equation part 23 (October 2021).