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8.7: Rational Equations - Mathematics


Rational Equations

Rational Equations

When one rational expression is set equal to another rational expression, a rational equation results.

Some examples of rational equations are the following (except for number 5):

Example (PageIndex{1})

(dfrac{3x}{4} = dfrac{15}{2})

Example (PageIndex{2})

(dfrac{x+1}{x-2} = dfrac{x-7}{x-3})

Example (PageIndex{3})

(dfrac{5a}{2} = 10)

Example (PageIndex{4})

(dfrac{3}{x} + dfrac{x-3}{x+1} = dfrac{6}{5x})

Example (PageIndex{5})

(dfrac{x-6}{x+1}) is a rational expression, not a rational equation.

The Logic Behind The Process

It seems most reasonable that an equation without any fractions would be easier to solve than an equation with fractions. Our goal, then, is to convert any rational equation to an equation that contains no fractions. This is easily done.

To develop this method, let’s consider the rational equation

(dfrac{1}{6} + dfrac{x}{4} = dfrac{17}{12})

The LCD is 12. We know that we can multiply both sides of an equation by the same nonzero quantity, so we’ll multiply both sides by the LCD, 12.

(12(dfrac{1}{6} + dfrac{x}{4}) = 12 cdot dfrac{17}{12})

Now distribute 12 to each term on the left side using the distributive property.

(12 cdot dfrac{1}{6} + 12 cdot dfrac{x}{4} = 12 cdot dfrac{17}{12})

Now divide to eliminate all denominators.

(egin{array}{flushleft}
2 cdot 1 + 3 cdot x &= 17
2 + 3x &= 17
end{array})

Now there are no more fractions, and we can solve this equation using our previous techniques to obtain 5 as the solution.

The Process

We have cleared the equation of fractions by multiplying both sides by the LCD. This development generates the following rule.

Clearing an Equation of Fractions

To clear an equation of fractions, multiply both sides of the equation by the LCD.

When multiplying both sides of the equation by the LCD, we use the distributive property to distribute the LCD to each term. This means we can simplify the above rule.

Clearing an Equation of Fractions

To clear an equation of fractions, multiply every term on both sides of the equation by the LCD.

The complete method for solving a rational equation is

1. Determine all the values that must be excluded from consideration by finding the values that will produce zero in the denominator (and thus, division by zero). These excluded values are not in the domain of the equation and are called nondomain values.

2. Clear the equation of fractions by multiplying every term by the LCD.

3. Solve this nonfractional equation for the variable. Check to see if any of these potential solutions are excluded values.

4. Check the solution by substitution.

Extraneous Solutions

Extraneous Solutions

Potential solutions that have been excluded because they make an expression undefined (or produce a false statement for an equation) are called extraneous solutions. Extraneous solutions are discarded. If there are no other potential solutions, the equation has no solution.

Sample Set A

Solve the following rational equations.

Example (PageIndex{6})

(egin{array}{flushleft}
dfrac{3x}{4} &= dfrac{15}{2} & ext{ Since the denominators are constants, there are no excluded values.}
&& ext{ No values must be excluded. The LCD is 4. Multiply each term by 4}
4 cdot dfrac{3x}{4} &= 4 cdot dfrac{15}{2}
cancel{4} cdot dfrac{3x}{cancel{4}} &= _{cancel{4}}^{2} cdot dfrac{15}{cancel{2}}
3x &= 2 cdot 15
3x &= 30
x &= 10 & 10 ext{ is not an excluded value. Check it as a solution}.
end{array})

Check:

(egin{array}{flushleft}
dfrac{3x}{4} &= dfrac{15}{2}
dfrac{3(10)}{4} &= dfrac{15}{2} & ext{ Is this correct? }
dfrac{30}{4} &= dfrac{15}{2} & ext{ Is this correct? }
dfrac{15}{2} &= dfrac{15}{2} & ext{ Yes, this is correct }
end{array})

Example (PageIndex{7})

(egin{aligned}
dfrac{4}{x-1} &= dfrac{2}{x+6} & 1 ext{ and } -6 ext{ are nondomain values. Exclude them from the solution}
&& ext{The LCD is } (x-1)(x+6) ext{ Multiply every term by the LCD }
(x-1)(x+6) cdot dfrac{4}{x-1} &= (x-1)(x+6) cdot dfrac{2}{x+6}
cancel{(x-1)}(x+6) cdot dfrac{4}{cancel{x-1}} &= (x-1)cancel{(x+6)} cdot dfrac{2}{cancel{x+6}}
4(x+6) &= 2(x-1) & ext{ Solve this nonfractional equation }
4x + 24 &= 2x - 2
2x &= -26
x &= -13 & -13 ext{ is not an excluded value. Check it as a solution}
end{aligned})

Check:

(egin{array}{flushleft}
dfrac{4}{x-1} &= dfrac{2}{x+6}
dfrac{4}{-13-1} &= dfrac{2}{-13 + 6} & ext{ Is this correct?}
dfrac{4}{-14} &= dfrac{2}{-7} & ext{ Is this correct?}
dfrac{2}{-7} &= dfrac{2}{-13 + 6} & ext{ Yes, this is correct }
end{array})

(-13) is the solution.

Example (PageIndex{8})

(egin{array}{flushleft}
dfrac{4a}{a-4} &= 2 + dfrac{16}{a-4}. & 4 ext{ is a nondomain value. Exclude it from consideration}
&& ext{ The LCD is } a-4 ext{. Multiply every term by } a-4
(a-4) cdot dfrac{4a}{a-4} &= 2(a-4) + (a-4) cdot dfrac{16}{a-4}
cancel{(a-4)} cdot dfrac{4a}{cancel{a-4}} &= 2(a-4) + cancel{(a-4)} cdot dfrac{16}{cancel{a-4}}
4a &= 2(a-4) + 16 & ext{ Solve this nonfractional equation }
4a &= 2a - 8 + 16
4a &= 2a + 8
2a &= 8
a &= 4
end{array})

This value, (a = 4), has been excluded from consideration. It is not to be considered as a solution. It is extraneous. As there are no other potential solutions to consider, we conclude that this equation has no solution.

Practice Set A

Solve the following rational equations.

Practice Problem (PageIndex{1})

(dfrac{2x}{5} = dfrac{x-14}{6})

Answer

(x=−10)

Practice Problem (PageIndex{2})

(dfrac{3a}{a-1} = dfrac{3a + 8})

Answer

(a=−2)

Practice Problem (PageIndex{3})

(dfrac{3}{y-3} + 2 = dfrac{y}{y-3})

Answer

(y = 3) is extraneous, so no solution.

Sample Set B

Solve the following rational equations.

Example (PageIndex{9})

(egin{array}{flushleft}
dfrac{3}{x} + dfrac{4x}{x-1} &= dfrac{4x^2 + x + 5}{x^2 - x} & ext{ Factor all denominators to find any excluded values and the LCD }
&& ext{ Nondomain values are } 0 ext{ and } 1. ext{ Exclude them from consideration. }
dfrac{3}{x} + dfrac{4x}{x-1} &= dfrac{4x^2 + x + 5}{x(x-1)} & ext{ The LCD is } x(x-1) ext{. Multiply each term by } x(x-1) ext{ and simplify }
end{array})
(cancel{x}(x-1) cdot dfrac{3}{cancel{x}} + x(cancel{x-1}) cdot dfrac{4x}{cancel{x-1}} = cancel{x(x-1)} cdot dfrac{4x^2 + x + 5}{cancel{x(x-1)}}).
(egin{array}{flushleft}
3(x-1) + 4x cdot x &= 4x^2 + x + 5 & ext{ Solve this nonfractional equation to obtain the potential solutions }
3x - 3 + 4x^2 &= 4x^2 + x + 5
3x - 3 &= x + 5
2x &= 8
x &= 4 & 4 ext{ is not an excluded value. Check it as a solution }
end{array})

Check:

(egin{array}{flushleft}
dfrac{3}{x} + dfrac{4x}{x-1} &= dfrac{4x^2 + x + 5}{x^2 - x}
dfrac{3}{4} + dfrac{4 cdot 4}{4-1} &= dfrac{4 cdot 4^2 + 4 + 5}{16 - 4} & ext{ Is this correct? }
dfrac{3}{4} + dfrac{16}{3} &= dfrac{64 + 4 + 5}{12} & ext{ Is this correct? }
dfrac{9}{12} + dfrac{64}{12} &= dfrac{73}{12} & ext{ Is this correct? }
dfrac{73}{12} &= dfrac{73}{12} & ext{ Yes, this is correct }
end{array})

(4) is the solution.

The zero-factor property can be used to solve certain types of rational equations. We studied the zero-factor property in Section 5.1, and you may remember that it states that if (a) and (b) are real numbers and that (a cdot b=0), then either or both (a=0) or (b=0).The zero-factor property is useful in solving the following rational equation.

Example (PageIndex{10})

(egin{array}{flushleft}
dfrac{3}{a^2} - dfrac{2}{a} &= 1 & ext{ Zero is an excluded value. }
&& ext{ The LCD is } a^2 ext{ Multiply each term by } a^2 ext{ and simplify }
cancel{a^2} cdot dfrac{3}{cancel{a^2}} - cancel{a^2} cdot dfrac{2}{cancel{a}} &= 1 cdot a^2
3-2a &= a^2 & ext{ Solve this nonfractional quadratic equation. Set it equal to zero }
0 &= a^2 + 2a - 3
0 &= (a+3)(a-1)
a&= - 3, a = 1 & ext{ Check these as solutions }
end{array})

Check:

(egin{array}{flushleft}
ext{If } a = -3: & dfrac{3}{(-3)^2} - dfrac{2}{-3} &= 1 & ext{ Is this correct? }
& dfrac{3}{9} + dfrac{2}{3} &= 1 & ext{ Is this correct? }
& dfrac{1}{3} + dfrac{2}{3} &= 1 & ext{ Is this correct? }
& 1 &= 1 & ext{ Yes, this is correct }
& a &= -3 & ext{ Checks and is a solution }
ext{If } a = 1: & dfrac{3}{(1)^2} - dfrac{2}{1} &= 1 & ext{ Is this correct? }
& dfrac{3}{1} - dfrac{2}{1} &= 1 & ext{ Is this correct? }
& 1 &= 1 & ext{ Yes, this is correct. }
& a &= 1 & ext{ Checks and is a solution }
end{array})

(-3) and (1) are the solutions.

Practice Set B

Practice Problem (PageIndex{4})

Solve the equation (dfrac{a+3}{a-2} = dfrac{a+1}{a-1})

Answer

(a = dfrac{1}{3})

Practice Problem (PageIndex{5})

Solve the equation (dfrac{1}{x-1} - dfrac{1}{x+1} = dfrac{2x}{x^2 - 1})

Answer

This equation has no solution. (x=1) is extraneous.

Section 7.6 Exercises

For the following problems, solve the rational equations.

Exercise (PageIndex{1})

(dfrac{32}{x} = dfrac{16}{3})

Answer

(x = 6)

Exercise (PageIndex{2})

(dfrac{54}{y} = dfrac{27}{4})

Exercise (PageIndex{3})

(dfrac{8}{y} = dfrac{2}{3})

Answer

(y=12)

Exercise (PageIndex{4})

(dfrac{x}{28} = dfrac{3}{7})

Exercise (PageIndex{5})

(dfrac{x + 1}{4} = dfrac{x-3}{2})

Answer

(x = 7)

Exercise (PageIndex{6})

(dfrac{a + 3}{6} = dfrac{a - 1}{4})

Exercise (PageIndex{7})

(dfrac{y-3}{6} = dfrac{y + 1}{4})

Answer

(y=−9)

Exercise (PageIndex{8})

(dfrac{x-7}{8} = dfrac{x+5}{6})

Exercise (PageIndex{9})

(dfrac{a + 6}{9} - dfrac{a-1}{6} = 0)

Answer

(a=15)

Exercise (PageIndex{10})

(dfrac{y + 11}{4} = dfrac{y + 8}{10})

Exercise (PageIndex{11})

(dfrac{b + 1}{2} + 6 = dfrac{b- 4}{3})

Answer

(b=−47)

Exercise (PageIndex{12})

(dfrac{m+3}{2} + 1 = dfrac{m-4}{5})

Exercise (PageIndex{13})

(dfrac{a - 6}{2} + 4 = -1)

Answer

(a=−4)

Exercise (PageIndex{14})

(dfrac{b + 11}{3} + 8 = 6)

Exercise (PageIndex{15})

(dfrac{y - 1}{y + 2} = dfrac{y + 3}{y - 2})

Answer

(y = -dfrac{1}{2})

Exercise (PageIndex{16})

(dfrac{x + 2}{x - 6} = dfrac{x - 1}{x + 2})

Exercise (PageIndex{17})

(dfrac{3m + 1}{2m} = dfrac{4}{3})

Answer

(m=−3)

Exercise (PageIndex{18})

(dfrac{2k + 7}{3k} = dfrac{5}{4})

Exercise (PageIndex{19})

(dfrac{4}{x + 2} = 1)

Answer

(x=2)

Exercise (PageIndex{20})

(dfrac{-6}{x - 3} = 1)

Exercise (PageIndex{21})

(dfrac{a}{3} + dfrac{10 + a}{4} = 6)

Answer

(a=6)

Exercise (PageIndex{22})

(dfrac{k + 17}{5} - dfrac{k}{2} = 2k)

Exercise (PageIndex{23})

(dfrac{2b + 1}{3b - 5} = dfrac{1}{4})

Answer

(b = -dfrac{9}{5})

Exercise (PageIndex{24})

(dfrac{-3a + 4}{2a - 7} = dfrac{-7}{9})

Exercise (PageIndex{25})

(dfrac{x}{x + 3} - dfrac{x}{x-2} = dfrac{10}{x^2 + x - 6})

Answer

(x=−2)

Exercise (PageIndex{26})

(dfrac{3y}{y-1} + dfrac{2y}{y-6} = dfrac{5y^2 - 15y + 20}{y^2 - 7y + 6})

Exercise (PageIndex{27})

(dfrac{4a}{a+2} - dfrac{3a}{a-1} = dfrac{a^2 - 8a - 4}{a^2 + a - 2})

Answer

(a=2)

Exercise (PageIndex{28})

(dfrac{3a - 7}{a-3} = dfrac{4a - 10}{a - 3})

Exercise (PageIndex{29})

(dfrac{2x - 5}{x - 6} = dfrac{x+1}{x-6})

Answer

No solution; 6 is an excluded value.

Exercise (PageIndex{30})

(dfrac{3}{x + 4} + dfrac{5}{x + 4} = dfrac{3}{x - 1})

Exercise (PageIndex{31})

(dfrac{2}{y + 2} + dfrac{8}{y + 2} = dfrac{9}{y + 3})

Answer

(y=−12)

Exercise (PageIndex{32})

(dfrac{4}{a^2 + 2a} = dfrac{3}{a^2 + a - 2})

Exercise (PageIndex{33})

(dfrac{2}{b(b+2)} = dfrac{3}{b^2 + 6b + 8})

Answer

(b=8)

Exercise (PageIndex{34})

(dfrac{x}{x-1} + dfrac{3x}{x-4} = dfrac{4x^2 - 8x + 1}{x^2 - 5x + 4})

Exercise (PageIndex{35})

(dfrac{4x}{x+2} - dfrac{x}{x+1} = dfrac{3x^2 + 4x + 4}{x^2 + 3x + 2})

Answer

no solution

Exercise (PageIndex{36})

(dfrac{2}{a-5} - dfrac{4a - 2}{a^2 - 6a + 5} = dfrac{-3}{a-1})

Exercise (PageIndex{37})

(dfrac{-1}{x+4} - dfrac{2}{x+1} = dfrac{4x + 19}{x^2 + 5x + 4})

Answer

No solution; (−4) is an excluded value.

Exercise (PageIndex{38})

(dfrac{2}{x^2} + dfrac{1}{x} = 1)

Exercise (PageIndex{39})

(dfrac{6}{y^2} - dfrac{5}{y} = 1)

Answer

(y=−6, 1)

Exercise (PageIndex{40})

(dfrac{12}{a^2} - dfrac{4}{a} = 1)

Exercise (PageIndex{41})

(dfrac{20}{x^2} - dfrac{1}{x} = 1)

Answer

(x=4, −5)

Exercise (PageIndex{42})

(dfrac{12}{y} + dfrac{12}{y^2} = -3)

Exercise (PageIndex{43})

(dfrac{16}{b^2} + dfrac{12}{b} = 4)

Answer

(y=4,−1)

Exercise (PageIndex{44})

(dfrac{1}{x^2} = 1)

Exercise (PageIndex{45})

(dfrac{16}{y^2} = 1)

Answer

(y=4,−4)

Exercise (PageIndex{46})

(dfrac{25}{a^2} = 1)

Exercise (PageIndex{47})

(dfrac{36}{y^2} = 1)

Answer

(y=6,−6)

Exercise (PageIndex{48})

(dfrac{2}{x^2} + dfrac{3}{x} = 2)

Exercise (PageIndex{49})

(dfrac{2}{a^2} - dfrac{5}{a} = 3)

Answer

(a = dfrac{1}{3}, -2)

Exercise (PageIndex{50})

(dfrac{2}{x^2} + dfrac{7}{x} = -6)

Exercise (PageIndex{51})

(dfrac{4}{a^2} + dfrac{9}{a} = 9)

Answer

(a = -dfrac{1}{3}, dfrac{4}{3})

Exercise (PageIndex{52})

(dfrac{2}{x} = dfrac{3}{x+2} + 1)

Exercise (PageIndex{53})

(dfrac{1}{x} = dfrac{2}{x+4} - dfrac{3}{2})

Answer

(x = -dfrac{4}{3}, -2)

Exercise (PageIndex{54})

(dfrac{4}{m} - dfrac{5}{m-3} = 7)

Exercise (PageIndex{55})

(dfrac{6}{a + 1} - dfrac{2}{a-2} = 5)

Answer

(a = dfrac{4}{5}, 1)

For the following problems, solve each literal equation for the designated letter.

Exercise (PageIndex{56})

(V = dfrac{GMm}{D}) for (D)

Exercise (PageIndex{57})

(PV = nrt) for (n).

Answer

(n = dfrac{PV}{rt})

Exercise (PageIndex{58})

(E = mc^2) for (m)

Exercise (PageIndex{59})

(P = 2(1 + w)) for (w).

Answer

(W = dfrac{P - 2}{2})

Exercise (PageIndex{60})

(A = dfrac{1}{2}h(b + B)) for (B).

Exercise (PageIndex{61})

(A = P(1 + rt)) for (r).

Answer

(r = dfrac{A - P}{Pt})

Exercise (PageIndex{62})

(z = dfrac{x-hat{x}}{s}) for (hat{x})

Exercise (PageIndex{63})

(F=dfrac{S_{x}^{2}}{S_{y}^{2}} ext { for } S_{y}^{2})

Answer

(S_{y}^{2}=dfrac{S_{x}^{2}}{F})

Exercise (PageIndex{64})

(dfrac{1}{R} = dfrac{1}{E} + dfrac{1}{F}) for (F).

Exercise (PageIndex{65})

(K = dfrac{1}{2}h(s_1 + s_2)) for (s_2).

Answer

(S_{2}=dfrac{2 K}{h}-S_{1} ext { or } dfrac{2 K-h S_{1}}{h})

Exercise (PageIndex{66})

(Q = dfrac{2mn}{s + t}) for (s).

Exercise (PageIndex{67})

(V = dfrac{1}{6}pi(3a^2 + h^2)) for (h^2).

Answer

(h_{2}=dfrac{6 V-3 pi a^{2}}{pi})

Exercise (PageIndex{68})

(I = dfrac{E}{R + r}) for (R).

Exercises For Review

Exercise (PageIndex{69})

Write ((4x^3y^{-4})^{-2}) so that only positive exponents appear.

Answer

(dfrac{y^8}{16x^6})

Exercise (PageIndex{70})

Factor (x^4 - 16)

Exercise (PageIndex{71})

Supply the missing word. An slope of a line is a measure of the _____ of the line.

Answer

steepness

Exercise (PageIndex{72})

Find the product (dfrac{x^{2}-3 x+2}{x^{2}-x-12} cdot dfrac{x^{2}+6 x+9}{x^{2}+x-2} cdot dfrac{x^{2}-6 x+8}{x^{2}+x-6})

Exercise (PageIndex{73})

Find the sum. (dfrac{2x}{x+1} + dfrac{1}{x-3})

Answer

(dfrac{2x^2 - 5x + 1}{(x+1)(x-3)})


Mathematics Solutions for Class 8 Math Chapter 1 - Rational And Irrational Numbers

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Page No 2:

Question 1:

Show the following numbers on a number line. Draw a separate number line for each example.

Answer:


1   3 2 , 5 2 , - 3 2 can be represented on the number line as follows.


2   7 5 , - 2 5 , - 4 5 can be represented on the number line as follows.


3   - 5 8 , 11 8 can be represented on the number line as follows.


4   13 10 , - 17 10 can be represented on the number line as follows.

Page No 2:

Question 2:

Observe the number line and answer the questions.

(1) Which number is indicated by point B?
(2) Which point indicates the number 1 3 4 ?
(3) State whether the statement, 'the point D denotes the number 5 2 , is true of false.

Answer:

(1) We observe that each unit on the number line is divided into 4 equal parts.
Now, B is the tenth point on the left of 0.
So, B indicates - 10 4 on the number line .
2   1 3 4 = 7 4 = 7 × 1 4
So, the seventh point on the right of 0 is C that indicates 1 3 4 on the number line.
(3) The point D is the tenth point on the right of 0. So, D indicates 10 4 on the number line.
Now, 10 4   =   5 2
So, D denotes 5 2 on the number line. Hence, the given statement is true.

Page No 3:

Question 1:

Compare the following numbers.
(1) &minus7, &minus2

Answer:

(2) We know that, a negative number is always less than 0.

(3) We know that, a positive number is always greater than 0.

7   15 12 = 15 × 4 12 × 4 = 60 48   7 16 = 7 × 3 16 × 3 = 21 48
Now, 60 48 > 21 48
&there4 15 12 > 7 16 .

(8) Let us first compare 25 8 and 9 4 .
25 8 = 25 × 1 8 × 1 = 25 8   9 4 = 9 × 2 4 × 2 = 18 8
Now, 25 8 > 18 8
&there4 25 8 > 9 4
&there4 - 25 8 < - 9 4 .

9   12 15 = 12 × 1 15 × 1 = 12 15   3 5 = 3 × 3 5 × 3 = 9 15
Now, 12 15 > 9 15
&there4 12 15 > 3 5 .

(10) Let us first compare 7 11 and 3 4 .
7 11 = 7 × 4 11 × 4 = 28 44   3 4 = 3 × 11 4 × 11 = 33 44
Now, 28 44 < 33 44
&there4 7 11 < 3 4
&there4 - 7 11 > - 3 4 .

Page No 4:

Question 1:

Write the following rational numbers in decimal form.
(1) 9 37

Answer:

(1) The given number is 9 37 .

&there4 9 37 = 0.243243. = 0 . 243
The decimal form of 9 37 is 0 . 243 .

(2) The given number is 18 42 .

&there4 18 42 = 0.428571428571. = 0 . 428571
The decimal form of 18 42 is 0 . 428571 .

(3) The given number is 9 14 .

&there4 9 14 = 0.6428571428571. = 0 . 6 428571
The decimal form of 9 14 is 0 . 6 428571 .

(4) The given number is - 103 5 .

&there4 103 5 = 20.6
The decimal form of - 103 5 is &minus20.6.

(5) The given number is - 11 13 .

&there4 11 13 = 0.846153846153. = 0 . 846153
The decimal form of - 11 13 is - 0 . 846153 .

Page No 5:

Question 1:

The number 2 is shown on a number line. Steps are given to show 3 on the number line using 2 . Fill in the boxes properly and complete the activity.

∙ The point Q on the number line shows the number .
∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
∙ Right angled ∆ ORQ is obtained by drawing seg OR.
l(OQ) = 2 , l(QR) = 1

Answer:


∙ The point Q on the number line shows the number 2 .
∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
∙ Right angled ∆ ORQ is obtained by drawing seg OR.
l(OQ) = 2 , l(QR) = 1
&there4 by Pythagoras theorem,
[l(OR)] 2 = [l(OQ)] 2 + [l(QR)] 2
= 2 2 + 1 2 = 2 + 1
= 3
&there4 l(OR) = 3
Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number line 3 .

Page No 6:

Question 2:

Show the number 5 on the number line.

Answer:


Draw a number line as shown in the figure. Let the point O represent 0 and point Q represent 2. Draw a perpendicular QR at Q on the number line such that QR = 1 unit. Join OR. Now, ∆OQR is a right angled triangle.
By Pythagoras theorem, we have
OR 2 = OQ 2 + QR 2
= (2) 2 + (1) 2
= 4 + 1
= 5
&there4 OR = 5
Taking O as the centre and radius OR = 5 , draw an arc cutting the number line at C.
Clearly, OC = OR = 5 .
Hence, C represents 5 on the number line.

Page No 6:

Question 3:

Show the number 7 on the number line.

Answer:


Draw a number line as shown in the figure and mark the points O, A and B on it such that OA = AB = 1 unit. The point O represents 0 and B represents 2. At B, draw CB perpendicular on the number line such that BC = 1 unit. Join OC. Now, ∆OBC is a right angled triangle.
In ∆OBC, by Pythagoras theorem
(OC) 2 = (OB) 2 + (BC) 2
= (2) 2 + (1) 2
= 4 + 1
= 5
&there4 OC = 5
Taking O as centre and radius OC = 5 , draw an arc cutting the number line at D.
Clearly, OC = OD = 5
At D, draw ED perpendicular on the number line such that ED = 1 unit. Join OE. Now, ∆ODE is a right angled triangle.
In ∆ODE, by Pythagoras theorem
(OE) 2 = (OD) 2 + (DE) 2
= ( 5 ) 2 + (1) 2
= 5 + 1
= 6
&there4 OE = 6
Taking O as centre and radius OE = 6 , draw an arc cutting the number line at F.
Clearly, OE = OF = 6
At F, draw GF perpendicular on the number line such that GF = 1 unit. Join OG. Now, ∆OFG is a right angled triangle.
In ∆OFG, by Pythagoras theorem
(OG) 2 = (OF) 2 + (FG) 2
= ( 6 ) 2 + (1) 2
= 6 + 1
= 7
&there4 OG = 7
Taking O as centre and radius OG = 7 , draw an arc cutting the number line at H.
Clearly, OG = OH = 7
Hence, H represents 7 on the number line.


8.7: Rational Equations - Mathematics

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The behavior of this method follows IEEE Standard 754, section 4. This kind of rounding is sometimes called rounding toward negative infinity.

Returns the largest integral value less than or equal to the specified double-precision floating-point number.

Parameters

A double-precision floating-point number.

Returns

The largest integral value less than or equal to d . If d is equal to NaN, NegativeInfinity, or PositiveInfinity, that value is returned.

Examples

The following example illustrates the Math.Floor(Double) method and contrasts it with the Ceiling(Double) method.

Remarks

The behavior of this method follows IEEE Standard 754, section 4. This kind of rounding is sometimes called rounding toward negative infinity. In other words, if d is positive, any fractional component is truncated. If d is negative, the presence of any fractional component causes it to be rounded to the smaller integer. The operation of this method differs from the Ceiling method, which supports rounding toward positive infinity.

Starting with Visual Basic 15.8, the performance of Double-to-integer conversion is optimized if you pass the value returned by the Floor method to the any of the integral conversion functions, or if the Double value returned by Floor is automatically converted to an integer with Option Strict set to Off. This optimization allows code to run faster -- up to twice as fast for code that does a large number of conversions to integer types. The following example illustrates such optimized conversions:


Homeschool Math Blog

Math teaching ideas, links, worksheets, reviews, articles, news, Math Mammoth stuff, and more - anything that helps YOU to teach math.

Saxon Math is not for everyone

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People sometimes ask me of my opinion or review of Saxon math. What I've written here applies in particular to Saxon Math's high school courses and middle grade levels. (The grades K-3 are by a different author and are quite different more on that below.)

Saxon Math uses an "incremental approach" where math concepts are studied in little pieces over several lessons, and those lessons are strawed over a long period of time, intermixed with lessons about other topics.

In other words, if one lesson is on some particular topic (say, percentages or inequalities), it's almost guaranteed that the NEXT lesson is NOT on that topic. It jumps around from topic to topic constantly, and this is by design.

Saxon's method also includes a feature where after a lesson is taught, there are very few practice problems about the topic of the lesson. Most of the problems are mixed review problems, and they practice concepts from earlier lessons, not the concept or skill of the lesson.

This PDF file contains three entire lessons from Saxon Algebra 1 so you can see for yourself how each lesson mostly has exercises about OTHER topics.

This type of arrangement helps students to MEMORIZE the content, since they get to practice any particular topic for quite a few days (though just a few problems per day). The downside is, it encourages many students to simply use rote memorization, and it does not guarantee nor promote conceptual understanding. Also, this approach can be very confusing to some students, and worse yet, turns some of them to math haters.

The instruction in the lessons seems adequate as do the exercises and problems I don't see big problems there. And I know many people like Saxon Math and many students have done well with it. It's possible to learn math well using Saxon Math, no doubt about it. But Saxon math can also be be disastrous. Personally I would rather see some sort of middle ground between constant review and the need to focus on new concepts.

If you do use Saxon, and notice that it's starting to turn your child against math as a subject, please consider other options. Also check that the child is not simply using rote memorization to get through, but that he/she does gain understanding of the CONCEPTS also.

The early levels aren't written by John Saxon but by Nancy Larson. She believes in conversation between the parent and the child, and that shows in the materials. The early courses are fully scripted, which I know some parents like and some don't. The suggested conversations seem good overall. Manipulatives are emphasized, which often is very good, but not all children need a lot of them. And in the early grade levels, the very tight spiraling works better than later on, so it's much less likely that the early grade levels of Saxon Math (K-3) would cause a child to start hating math because of the curriculum. Those levels of Saxon can work perfectly fine (depending on the child).

The middle grade levels are written by John Saxon and Stephen Hake. Here is where the tight spiraling can become a stumbling block.

I feel that homeschooling parents need to be aware that Saxon is not necessarily the "gold standard." It works for some children, and not for others, like all the other curricula out there. but over the years I've gotten the impression that for some reason (maybe because of its popularity), many parents, especially those starting to home school, tend to think of Saxon a bit higher than necessary. They just automatically choose Saxon because "everyone uses it." So then there is more of a chance of real damage being done than with most any other math curriculum. It seems that people tend to have the mindset that it should work (even when it isn't working) since so many other people use it.

So, I'd like to bring some awareness to the potential pitfalls of Saxon Math from 4th grade onward, and the main potential problem is the tight spiraling. The instruction is good and the exercises/activities are good but the organization of the materials can cause trouble.

But no matter what curriculum you use, remember that the TEACHER (you) is the most important part of the whole experience! You can try to adapt the curriculum, such as do less problems or use it in a different order than the table of contents indicates (though with Saxon that would be challenging). The TEACHER is what can make the biggest difference in how and what the student learns. Don't be a slave to any curriculum, not to Math Mammoth either!


I'm not the only one who feels critical of Saxon Math's approach. I've read plenty of other opinions along the same lines. here are two I found on Amazon:

"I am a math tutor, and I have to say that this Algebra book was not useful at all! The organization of the concepts is illogical, the 'progressive' methodology is confusing and the practice problems are badly conceived. The girl I tutor had already forgotten the lesson she had just learned within a few days because instead of providing enough problems to enforce the lesson taught, the creators of this book decided to review previously learned lessons for the majority of their exercises. I found myself having to re-teach lessons every day." (R.U. Kidding)

"Tutoring higher math, from Algebra I through Calculus and Differential Equations, for nearly 30 years, I feel qualified to blast this book! I tried to use it with home-schooled students. They were becoming increasingly confused, so I changed their books quickly. The 'Saxon' students that I have tutored in higher level math, needed remedial tutoring in Algebra I concepts before I could move on to their current grade level subject matter. Although Saxon is quite revered among homeschoolers, I cannot see why. The topics are not clearly defined. The same concepts are repeated over and over, lesson after lesson. If one needs to look up information, they cannot be sure which section of the book to read. The flow of concepts is erratic, at best. It is, in my opinion, detrimental to homeschoolers, who are quite often very intelligent, and deserve a much clearer, more concise text, that covers more topics associated with classical Algebra I." (Denise Sipe)

Read how professor Hung-Hsi Wu has worded it (emphases and the additional note are mine):

"But I think that what perhaps disturbs me the most about Saxon is to read through it, I myself do not get the feeling that I am reading something that when that the children use it they would even have a remotely correct impression of what mathematics is about. It is extremely good at promoting procedural accuracy. And what David says about building everything up in small increments, that's correct, but the great pedagogy is devoted, is used, to serve only one purpose, which is to make sure that the procedures get memorized, get used correctly. And you would get the feeling that-I think of it as a logical analogy-you can see the skeleton presented with quite a bit of clarity, but you never see any methods, your never see any flesh, nothing-no connective tissue, you only see the bare stuff.

A little bit of this is okay, but when you read through a whole volume of it really I am very, very, uneasy. There are lots of things in it that I admire, but something that is so one-sided-you think once more about yourself and you think about what happens if this thing gets adopted. There might be lots and lots of children using it. And suppose that hundreds of thousands of students are using this book and they go through four years of it. Would you be willing to face the end result? That here are hundreds of thousands of students thinking that mathematics is basically a collection of techniques."


8.7: Rational Equations - Mathematics

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Roman Vershynin, Department of Mathematics, UC Irvine

Email: rvershyn "at" uci "dot" edu

Office hours: MW 2:10 - 3:00pm in 540D Rowland Hall

Teaching Assistant

Boya Liu, Department of Mathematics, UC Irvine

Email: boyaliu1129 "at" gmail "dot" com

Office hours: Tu 11:00am - 1:00pm, W 10:00 - 11:00am in 250A Rowland Hall

When & Where

Lectures: MWF 12:00-12:50pm (Section 44779) and 1:00-1:50pm (Section 44775) in SH 174

Discussion: TuTh 1:00-1:50pm in SSTR 103 (Section 44780) and 10:00-10:50am in SST 120 (Section 44776)

Description, Prerequisites & Textbook

Course description: Introduction to real analysis, including convergence of sequence, infinite series, differentiation and integration, and sequences of functions. Students are expected to do proofs. Chapters 1-3 (except 3.19, 3.20) will be covered.

Prerequisites: Prerequisites: (MATH 2B or AP Calculus BC) and (MATH 2D or MATH H2D) and (MATH 3A or MATH H3A) and MATH 13. AP Calculus BC with a minimum score of 4. MATH 13 with a grade of C or better.

Textbook: K. Ross, Elementary Analysis, second edition.

Grading

The course grade will be determined as follows:

  • Homework: 10%. One homework with the lowest score will be dropped. Solutions will be collected every Thursday. Late homework will not be accepted. You are welcome and encouraged to form study groups and discuss homework with other students, but you must write your solutions individually.
  • Midterm Exam 1: 25%, Wednesday, October 24, in class. Covers everything covered in class up to, and including, October 17.
  • Midterm Exam 2: 25%, Monday, November 19, in class. Covers everything covered in class up to, and including, November 9.
  • Final Exam: 40%, Wednesday, December 12, 1:30--3:30pm, in ICS 174.

There will be no make-up for the exams for any reason. A missed midterm exam counts as zero points, with the following exception. If you miss a midterm exam due to a documented medical or family emergency, the exam's weight will be added to the weight of the final exam.


Linear Equation Problems

Value of the unknown quantity for which from given equation we get true numerical equality is called root of that equation. Two equations are called equivalent when the multitudes of their roots match, the roots of the first equation are also roots of the second and vice versa. The following rules are valid:
1. If in given equation one expression is substituted with another identity one, we get equation equivalent to the given.
2. If in given equation some expression is transferred from one side to the other with contrary sign, we get equation equivalent to the given.
3. If we multiply or divide both sides of given equation with the same number, different from zero, we get equation equivalent to the given.
Equation of the kind $ax + b = 0$, where $a, b$ are given numbers is called simple equation in reference to the unknown quantity $x$.

Problem 1 Solve the equation:
A) 16x + 10 – 32 = 35 – 10x - 5
B) $y + frac<3> <2y>+ 25 = frac<1><2>y + frac<3><4>y – frac<5><2>y + y + 37$
C) 7u – 9 – 3u + 5 = 11u – 6 – 4u

A)We do some of the makred actions and we get
16x – 22 = 30 – 10x
After using rule 2 we find 16x + 10x = 30 + 22
After doing the addition 26x = 52
We find unknown multiplier by dividing the product by the other multiplier.
That is why $x = frac<52><26>$
Therefore x = 2

B) By analogy with A) we find:
$yleft(1 + frac<3><2> ight) + 25 = yleft(frac<1> <2>+ frac<3> <4>– frac<5> <2>+ 1 ight) + 37 Leftrightarrow$
$frac<5><2>y + 25 = -frac<1><4>y + 37 Leftrightarrow frac<5><2>y + frac<1><4>y = 37 - 25 Leftrightarrow$
$frac<11><4>y = 12 Leftrightarrow y = frac<12.4> <11>Leftrightarrow y = frac<48><11>$

C) 4u – 4 = 7u – 6 6 – 4 = 7u – 4u 2 = 3u $u = frac<2><3>$

Problem 2 Solve the equation :
A) 7(3x – 6) + 5(x - 3) - 2(x - 7) = 5
B) (x -3)(x + 4) - 2(3x - 2) = (x - 4) 2
C) (x + 1) 3 – (x - 1) 3 = 6(x 2 + x + 1)

A) 21x - 42 + 5x - 15 - 2x + 14 = 5
21x + 5x - 2x = 5 + 42 + 15 - 14
24x = 48 x = 2

B) x 2 + 4x - 3x - 12 - 6x + 4 = x 2 - 8x + 16
x 2 - 5x – x 2 + 8x = 16 + 12 – 4
3x = 24 x = 8

C) x 3 + 3x 2 + 3x + 1 – (x 3 - 3x 2 + 3x - 1) = 6x 2 + 6x + 6
x 3 + 3x 2 + 3x + 1 – x 3 + 3x 2 + 1 = 6x 2 + 6x + 6
2 = 6x + 6 6x = -4 $x = -frac<2><3>$

A) $frac<5x - 4> <2>– frac<0.5x + 1> <3>Leftrightarrow$
3(5x - 4) = 2(0.5x + 1)
15x - 12 = x + 2
15x – x = 12 + 2
14x = 14 x = 1

B) $1 – frac <5>= frac<3(1 - x)><3>Leftrightarrow$
$1 –frac <5>= 1 – xLeftrightarrow$
-x + 3 = - 5x
5x – x = - 3
$x = -frac<3><4>$

C) $frac<3(x - 1)> <2>+ frac<2(x + 2)> <4>= frac<3x + 4.5> <5>Leftrightarrow$
$frac<2(x + 1) - 3(2x + 5)> <6>= - 3 Leftrightarrow$
$frac<2x + 2 - 6x -15> <6>= - 3 Leftrightarrow$
-4x - 13 = -18
-4x = -18 + 13
-4x = -5 $x = frac<5><4>$

D) We reduce to common denominator, which for 2, 4 and 5 is 20
$frac<3(x - 1)> <2>+ frac<2(x + 2)> <4>= frac<3x + 4.5> <5>Leftrightarrow$
30(x - 1) + 10(x + 2) = 4(3x + 4.5)
30x - 30 + 10x + 20 = 12x + 18
40x - 12x = 18 + 10
28x = 28 x = 1

Problem 4 Proof that every value of the unknown quantity is root of the equation:
A) 7x - 13 = - 13 + 7x
B) $(frac<1> <2>– x)^2 – (frac<1> <2>+ x)^2 = -2x$
C) 3x - 3x = 26 - 2(7 + 6)
D) $frac<-3x + 4x^2> <5>= (0.8x - 0.6)x$

Solution: For one simple equation with unknown quantity x every x is a solution, if it is reduced to the following equivalent equation 0.x = 0 or it transforms into identity a = a . Actually, in the left any value of x , when we multiply it with zero, will obtain zero, i.e. the right side or the value of x won’t influence the right or left side of the identity.

A) 7x - 7x = -13 + 13 0.x = 0 every x is a solution.

B) $frac<1> <4>- x + x^2 –(frac<1> <4>+ x + x^2) = - 2x$
$frac<1> <4>- x + x^2 -frac<1> <4>– x – x^2 = - 2x$
-2x = -2x
-2x + 2x = 0
0.x = 0 Therefore every x is a solution.

C) 0.x = 26 - 2.13
0.x = 26 – 26
0.x = 0 every x is a solution.

D) -3x + 4x 2 = 5(0.8x - 0.6)x
-3x + 4x 2 = (4x - 3)x
-3x + 4x 2 = 4x 2 - 3x
Therefore every x is a solution.

Problem 5 Proof that the equation has no roots:
A) 0.x = 34
B) 5 - 3x = 7 - 3x
C) $frac <4>= frac<4>$
D) 2(3x - 1) – 3(2x + 1) = 6

A) For the left side we will get value 0 for every x and for the right side is 34, i.e. number different from 0. Therefore there is no such x to get a true numerical equality

B) 5 - 3x = 7 - 3x 3x - 3x = 7 - 5 0.x = 2 0 = 2, which is impossible for any x

C) $frac <4>= frac<4>$ x - 3 = x + 5 x – x = 5 + 3 0 = 8 => no solution

D) 2(3x - 1) - 3(2x + 1) = 6 6x - 2 - 6x - 3 = 6 0.x = 6 + 5 0 = 11 no solution.

Problem 6 Solve the equation:
A) 2x 2 - 3(1 – x)(x + 2) + (x - 4)(1 - 5x) + 58 = 0
B) 3.(x + 1) 2 – (3x + 5).x = x + 3
C) x 2 – (x - 1).(x + 1) = 4
D) (x - 1).(x 2 + x + 1) = (x - 1) 3 + 3x(x - 1)
E) (3x - 1) 2 – x(15x + 7) = x(x + 1).(x - 1) – (x + 2) 3

A) 2x 2 - 3(x + 2 – x 2 - 2x) + x - 5x 2 - 4 + 20x + 58 = 0
2x 2 - 3x - 6 + 3x 2 + 6x + x - 5x 2 - 4 + 20x + 58 = 0
0.x 2 + 24x + 48 = 0
24x = - 48 x = -2

B) 3(x 2 + 2x + 1) - 3x 2 - 5x = 3x 2 + 6x + 3 - 3x 2 -5x = x + 3
(3 - 3)x 2 + (6 - 5).x – x = 3 - 3
0 = 0 => every x is a solution

C) x 2 – (x 2 -1) = 4
x 2 – x 2 + 1 = 4
0 = 3 => no solution

D) x 3 + x 2 + x – x 2 – x - 1 = x 3 - 3x 2 + 3x - 1 + 3x 2 - 3x
0 = 0 => every x is a solution

E) 9x 2 - 6x + 1 - 15x 2 - 7x = x 3 –x 2 + x 2 – x – x 3 - 6x 2 - 12x - 8
0 = 9 => no solution

Problem 7 Solve the equation:
A) $frac<6x - 1> <5>- frac<1 - 2x> <2>= frac<12x + 49><10>$
B) $frac <2>+ frac<2x - 2> <4>= frac<7x - 6><3>$

A)We reduce to common denominator and we get:
12x - 2 - 5 +10x = 12x + 49
22x - 12x = 49 + 7
10x = 56 x = 5.6

Problem 8 The function f(x) = x + 4 is given. Solve the equation:
$frac<3f(x - 2)> + 4 = f(2x + 1)$

We calculate f(0), f(x -2), f(2x +1), namely f(0) = 0 + 4 = 4
f(x - 2) = x - 2 + 4 = x + 2
f(2x + 1) = 2x + 1 + 4 = 2x + 5 The equation gets this look
$frac<3(x + 2)> <4>+ 4 = 2x + 5$
3(x + 2) +16 = 4(2x + 5)
3x + 6 +16 = 8x + 20
22 - 20 = 8x - 3x
2 = 5x x = 0.4

Problem 9 Solve the equation:
(2x - 1) 2 – x(10x + 1) = x(1 – x)(1 + x) – (2 – x) 3

(2x - 1) 2 – x(10x + 1) = x(1 – x)(1 + x) – (2 – x) 3
4x 2 - 4x + 1 -10x 2 – x = x – x 3 - 8 + 12x - 6x 2 + x 3
18x = 9 $x = frac<1><2>$

Problem 10 Solve the equation:
(2x + 3) 2 –x(1 + 2x)(1 - 2x) = (2x - 1) 2 + 4x 3 - 1

(2x + 3) 2 – x(1 + 2x)(1 - 2x) = (2x - 1) 2 + 4x 3 -1
4x 2 + 12x + 9 – x(1 - 4x 2 ) = 4x 2 - 4x + 1 + 4x 3 - 1
12x + 9 – x + 4x 3 = - 4x + 4x 3
15x = -9 $x = -frac<3><5>$

Problem 11 Solve the equation :
(2x - 1) 3 + 2x(2x - 3).(3 - 2x) – (3x - 1) 2 = 3x 2 - 2

We open the brackets by using the formulas for multiplication:
8x 3 - 3(2x) 2 .1 + 3.2x(1) 2 – 1 3 - 2x(2x - 3) 2 – (9x 2 - 6x + 1) = 3x 2 - 2
8x 3 - 12x 2 + 6x - 1 - 2x(4x 2 - 12x + 9) - 9x 2 + 6x - 1 = 3x 2 - 2
8x 3 - 21x 2 + 12x - 8x 3 + 24x 2 - 9x = 3x 2
3x 2 + 3x = 3x 2
3x = 0 x = 0

Problem 12 Solve the equation :
$left(2x - frac<1><2> ight)^2 – (2x - 3)(2x + 3) = x + frac<1><4>$

We use the formulas for multiplication, open the brackets and get:
$4x^2 - 2x + frac<1> <4>– (4x - 9) = x + frac<1><4>$
$4x^2 - 2x + frac<1> <4>- 4x^2 + 9 = x + frac<1><4>$
9 = x + 2x
9 = 3x x = 3

Problem 13 Proof that the two equations are equivalent:
A) $frac <2>+ frac <8>= frac<1.5x - 10><4>$ and $frac <2>– frac<5.5 - 0.5x> <3>= 1.5$
B) $x – frac<8x + 7> <6>+ frac <3>= -1left(frac<1><6> ight)$ and $2x – frac<6 – x> <3>- 2left(frac<1><3> ight)x = -2$

A) For the first equation we get consecutively:
4(x - 5) + x - 1 = 2(1.5x - 10)
4x - 20 + x - 1 = 3x - 20
5x – 3x = - 20 + 21
2x = 1 $x = frac<1><2>$,
for the second equation we have
3(x + 6) - 2(5.5 - 0.5y) = 6 . 1.5
3x + 18 - 11 + x = 9
4y = 9 - 7
$x = frac<2><4>$ $x = frac<1><2>$ Therefore the equations are equivalent.

B) Analogical to A) Try by yourself

Problem 14Solve the equation :
A) (2x + 1) 2 – x(1 - 2x)(1 + 2x) = (2x - 1) 2 + 4x 3 - 3
B) (2x - 1) 2 + (x - 2) 3 = x 2 (x - 2) + 8x - 7
C) (x + 2)(x 2 - 2x + 4) + x(1 – x)(1 + x) = x - 4
D) $frac<8x + 5> <4>– frac<1><2left[2 – frac<3 – x><3> ight]> = 2x + frac<5><6>$
E) $frac <3>– frac <4>= x – frac<1><3left[1 – frac<3 - 24x><8> ight]>$
F) $frac <5>– frac<(2x - 3)^2> <3>= frac<1><5>left[5 – frac<20x - 43x><3> ight]$

A) 4x 2 + 4x + 1 – x(1 - 4x 2 ) = 4x 2 - 4x + 1 + 4x 3 - 3
4x – x + 4x 3 = -4x + 4x 3 -3
3x + 4x = -3
7x = - 3 $x = -frac<3><7>$

B) 4x 2 - 4x + 1 + x 3 - 3x 2 .2 + 3x.2 2 - 8 = x 3 -2x 2 + 8x - 7
4x 2 - 6x 2 - 4x + 1 + 12x - 8 = - 2x 2 + 8x -7
-2x 2 + 8x - 7 = - 2x 2 + 8x - 7
0 = 0 => every x is a solution

C) x 3 + 2x 2 - 2x 2 - 4x + 4x + 8 + x(1 – x 2 ) = x - 4
x 3 + 8 + x – x 3 = x - 4
8 = - 4, which is impossible. Therefore the equation has no solution

D) $frac<8x + 5> <4>- 1 + frac<3 – x> <6>= 2x + frac<5> <6>Leftrightarrow$
3(8x + 5) - 12 + 2(3 – x) = 24x + 2.5
24x + 15 - 12 + 6 - 2x = 24x + 10
-2x = 10 - 9 $x = -frac<1><2>$

E) $frac <3>– frac <4>= x - frac<1> <3>+ frac<3 - 24x> <24>Leftrightarrow$
8x - 6(x + 3) = 24x - 8 + 3 - 24x
8x - 6x - 18 = -5
2x = 18 - 5
2x = 13 x = 6.5

F) $frac <5>– left[frac<2x - 3><3> ight]^2 = 1 – frac<20x^2 - 43x> <15>Leftrightarrow$
3x - 5(4x 2 -12x + 9) = 15 - 20x 2 + 43x
3x - 20x 2 + 60x - 45 = 15 - 20x 2 + 43x
63x - 43x = 15 + 45
20x = 60 x = 3


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Interventions

Student has difficulty getting started.

  • What are you trying to do?
  • What are the coordinates of the vertices of the original figure?
  • What axis are you reflecting the figure across?
  • What will you do first? Second?

Student has an incorrect solution.

  • Explain how you reflected the figure across the axis.
  • What does it mean to reflect a point across an axis?
  • How do you know that the point is reflected across the axis correctly?

Student has a correct solution.

  • What method did you use to reflect the figure across the axis?
  • Did you reach any conclusions about the coordinates of the reflected figure&rsquos vertices?

ELLs: In posing these questions, make sure that if the student involved is an ELL, your pace is adequate and you are providing ample wait time to allow for a thoughtful response.


Meets Expectations

The instructional materials reviewed for EdGems Math Grade 7 meet expectations for focus and coherence in Gateway 1. The instructional materials meet the expectations for focus by assessing grade-level content and devoting the large majority of class time to major work of the grade. The instructional materials meet expectations for coherence due to being consistent with the progressions in the standards and making connections within the grade.

Criterion 1a

The instructional materials reviewed for EdGems Math Grade 7 meet expectations for not assessing topics before the grade level in which the topic should be introduced. There are above grade-level assessment items that could be modified or omitted without impact on the underlying structure of the instructional materials.

Indicator 1a

The instructional materials reviewed for EdGems Math Grade 7 meet expectations for assessing grade-level content.

Each unit includes Form A and Form B Assessments as well as Tiered Assessments Form AT and Form BT, all of which include selected response and constructed response sections. Performance Tasks are also included with each unit. In addition, Gem Challenges are online, standards-based items for use after a standard has been addressed and are located after certain lessons.

Examples of grade-level assessments include:

  • Unit 2, Proportional Relationships, Form A, Part II, Problem 1: “Write two different ratios that would form a proportion with the ratio of 8/6.” (7.RP.1)
  • Unit 5, Products & Quotients of Rational Numbers, Form A, Part I, Problem 5: “What is the value of the expression below? 6(−1+5)−30 a. -66, b. -54, c. -6, d. 6” (7.NS.2a)
  • Unit 6, Algebraic Expressions, Form A, Part II, Problem 16: “Explain two different ways to simplify 3(1.2x − 6+2.1 x ). Show that both ways lead to the same simplified expression.” (7.EE.1)
  • Lesson 10.1, Probability, Online Gem Challenge 1, Problem 3: “Students in a math class will be randomly assigned a polygon for a class project. The only types of polygons being assigned are quadrilaterals, pentagons, hexagons, octagons and decagons. If there is an equal number of each type of polygon, what is the probability that the first polygon assigned will be a hexagon?” (7.SP.7)

There are above grade-level assessment items that could be modified or omitted without impact on the underlying structure of the instructional materials. These items include:

  • Unit 9, Part I, Problem 9: “The area of the base of a trapezoidal pyramid is $34ft^2$. The pyramid is 12 feet tall. What is the volume of the pyramid?” (8.G.9)
  • Unit 9, Part 1, Problem 10: “The perimeter of the base of a square pyramid is 12 yards. The height of the pyramid is 13.5 yards. What is the volume of the pyramid?” (8.G.9)
  • Unit 8, Part I, Problem 2: “What is the approximate area of the shaded sector? Use 3.14 for pi.” (G-C.5) Students are using a circle with a sector shaded. The angle within the sector is labeled as 100* and the radius is labeled as 3 cm.
  • Unit 8, Part II, Form A, Problem 4: “Determine if the following pair of triangles must be the same shape or not. Explain your reasoning.” (8.G.4)

Criterion 1b

The instructional materials reviewed for EdGems Math Grade 7 meet expectations for devoting the large majority of class time to the major work of the grade. The instructional materials spend approximately 73% of class time on the major work of the grade.

Indicator 1b

The instructional materials reviewed for EdGems Math Grade 7 meet expectations for spending a majority of instructional time on major work of the grade.

  • The number of units devoted to major work of the grade (including assessments and supporting work connected to the major work) is 6.5 out of 10, which is 65%.
  • The number of lessons devoted to major work of the grade (including supporting work connected to the major work) is 27.5 out of 43, which is approximately 64%.
  • The approximate number of days devoted to major work (including assessments and supporting work connected to the major work) is 102 out of 140, which is 73%.

A day-level analysis is most representative of the instructional materials because this perspective includes all connections to major work and follows the recommended pacing suggestions for addressing major work. As a result, approximately 73% of the instructional materials focus on major work of the grade.

Criterion 1c - 1f

The instructional materials reviewed for EdGems Math Grade 7 meet expectations for being coherent and consistent with the Standards. The instructional materials have supporting work that enhances focus and coherence simultaneously, are consistent with the progressions in the standards, and foster coherence through connections within the grade.

Indicator 1c

The instructional materials reviewed for EdGems Math Grade 7 meet expectations that supporting work enhances focus and coherence simultaneously by engaging students in the major work of the grade. Supporting standards and clusters are connected to major standards and clusters of the grade, and lessons address supporting standards while maintaining focus on the major work of the grade. Examples of supporting work being used to support the focus and coherence of the major work of the grade include:

  • Lesson 2.2 connects 7.G.6 and 7.RP.3 as students use facts about polygons to solve proportions. An example is, “Two squares have a scale of 1 : 8 1/2 . The perimeter of the larger square is 170 units. What is the side length of the smaller square?”
  • Lesson 8.4 connects 7.EE.3 and 7.G.6 as students write and solve equations to determine area and missing side lengths of polygons. An example is, “The length of a rectangle is 2.5 cm. The area is $20 cm^2$. What is the width of the rectangle?”
  • Lesson 10.2 connects 7.RP.2c and 7.SP.7 as students use proportions to predict outcomes using probability. For example, Example 2 states, “Last week in practice, Lou had 12 hits in 40 at-bats. Use experimental probability to predict how many hits he will have next week if he gets 30 at-bats.” The worked-out example describes how to set up a proportion to solve.
  • Lesson 10.3 connects 7.SP.8 and 7.NS.2 as students find the probability of events by multiplying rational numbers. An example is, “A shirt comes in three colors (blue, red and black) and can be either long-sleeved or short-sleeved. If you choose one shirt from a pile, what is the probability that it is a short-sleeved blue shirt?”

Indicator 1d

The instructional materials for EdGems Math Grade 7 partially meet expectations that the amount of content designated for one grade level is viable for one year.

As designed, the instructional materials can be completed in 110-144 days. If teachers followed the pacing guide, and used the minimal amount of days allocated, the materials would not be viable for a full school year. If teachers followed the pacing guide, and used the maximum amount of days allocated, the materials would be viable for a full school year. Considering the variability of instructional days, these materials partially meet expectations that the amount of content designated for one grade level is viable for one year.

The materials include ten units, containing 43 lessons. Lessons range in length from one to four days. Each unit includes lessons, assessments, and targeted interventions.

  • The Pacing Guide designates one lesson as 1-2 days, 22 lessons as 2-3 days, one lesson as 3-4 days, and 19 lessons as 2 days, leading to a total of 86-110 lesson days.
    • 1 lesson = 1 to 2 days
    • 22 lessons = 44 to 66 days
    • 1 lesson = 3 to 4 days
    • 19 lessons = 38 days

    Additionally, there is a discrepancy within the Grade 7 materials. Based on each unit overview page there is a range of 114-151 instructional days, with 86-110 days for lessons and 28-41 days for assessment. Based on the Scope and Sequence document, there is a range of 113-146 instructional days, with 86-110 days for lessons and 27-36 days for assessment. In addition, on the top of some of the Scope and Sequence documents within the units of Grade 7, it gives a range of 110-144 days, such as in Units 1 and 2, but Unit 10 gives a range of 121-160 days.

    Indicator 1e

    The instructional materials for EdGems Math Grade 7 meet expectations for being consistent with the progressions in the Standards. In general, the instructional materials clearly identify content from prior and future grade-levels and use it to support the progressions of the grade-level standards. In addition, the instructional materials give all students extensive work with grade-level problems.

    Each Unit Overview describes how the work of the unit is connected to previous grade level work, for example:

    • The introductory paragraph of the Unit 7 Overview, Solving Equations and Inequalities, states, “In Grade 6 CCSS, students solved one-step equations. In this unit, students will apply their understanding of balancing an equation to solving two-step equations. They will also use their skills of simplifying expressions to solve equations that include like terms or the Distributive Property. Students have previously used the inequality symbols to compare numbers and graph solutions to an inequality. Students will also combine that knowledge with the equation-solving process to solve inequalities and graph their solutions on a number line.”

    Each Unit Overview includes Learning Progression, and each Learning Progression includes statements identifying what students have learned in earlier grades and what students will learn in future grades, for example:

    Unit 6: Algebraic Expressions, In earlier grades, students have…

    • Evaluated expressions in which letters stand for numbers. (6.EE.2)
    • Applied properties of operations to generate equivalent expressions. (6.EE.3-4)
    • Used variables to represent numbers and write expressions for real-world problems. (6.EE.6)

    In future grades, students will…

    • Solve multi-step equations that require simplifying before solving. (8.EE.7)
    • Add, subtract and multiply polynomials. (A-APR)
    • Interpret expressions that represent a quantity in terms of its context. (A-SSE.1)

    In some units, the Unit Overview references connections to current grade level work that was addressed in prior units. Examples include:

    • Lesson 2.2, Problem Solving With Proportions, the Teaching Tips Section includes, “In this lesson, students utilize their knowledge of scale factors and scales from Lesson 1.4 and apply these scales using proportions.”
    • Lesson 5.3, Dividing Rational Numbers, “Students divided fractions when working with complex fractions in Unit 1. Make connections to that work to remind students about the process of dividing fractions by multiplying by the reciprocal.”

    The instructional materials present opportunities for students to engage with work with grade-level problems within each Student Lesson, Explore activity, Student Gem (online activities to provide practice with the content), Online Practice & Gem Challenge (only in some lessons), Exit Card, and Performance Task. For example:

    • In Lesson 5.4, students solve problems by identifying where to put parentheses in numerical expressions (7.NS.3). For example, “Insert one set of parentheses to make the equation true: Problem 31. 5 + 3 + 9 ÷ 3 = 9.”

    The materials include one example of off grade-level content that is not identified that distracts students from engaging with the grade-level standards:

    • In Lesson 8.6, students find the area of sectors of circles (G-C.5). Example 5 presents a diagram of a circle with a 115 degree shaded sector and states, “Find the area of the shaded sector in circle M. Round to the nearest hundredth.”

    Each unit includes a Parent Guide with Connecting Math Concepts, which includes, “Past math topics your child has learned that will be activated in this unit and Future math this unit prepares your child for.” For example, in Unit 6, Algebraic Expressions, “Past math topics your child has learned that will be activated in this unit evaluating expressions in which letters stand for numbers, applying properties of operations to generate equivalent expressions, and using variables to represent numbers and write expressions for real-world problems.” “Future math this unit prepares your child for solving multi-step equations that require simplifying before solving, adding, subtracting and multiplying polynomials, and interpreting expressions that represent a quantity in terms of its context.”

    Each Lesson Guide includes Teaching Tips, which often include connections from prior or future grades, for example:

    • Lesson 2.4, Proportional Relationships Equations, “In later grades (starting in Grade 8), students begin calling the constant of proportionality the slope of the line. You may want to connect to the concept of slope in this lesson as students are solidifying the idea that the constant of proportionality is the rate at which the function is increasing or decreasing. The larger the absolute value of the constant of proportionality, the steeper the line.”

    In each Lesson Guide, Warm Up includes problems noted with prior grade-level standards. For example:


    Watch the video: Solving Rational Equations (October 2021).