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5.8: Linear Equations in Two Variables


Solutions to Linear Equations in Two Variables

Solution to an Equation in Two Variables

We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. If the relationship is between two quantities, the equation will contain two variables. We say that an equation in two variables has a solution if an ordered pair of values can be found such that when these two values are substituted into the equation a true statement results. This is illustrated when we observe some solutions to the equation (y=2x+5).

(egin{array}{l}
x=4, y=13 ; quad ext { since } 13=2(4)+5 ext { is true }
x=1, y=7 ; quad ext { since } 7=2(1)+5 ext { is true. }
x=0, y=5 ; quad ext { since } 5=2(0)+5 ext { is true. }
x=-6, y=-7 ; quad ext { since }-7=2(-6)+5 ext { is true. }
end{array})

Ordered Pairs as Solutions

It is important to keep in mind that a solution to a linear equation in two variables is an ordered pair of values, one value for each variable. A solution is not completely known until the values of both variables are specified.

Independent and Dependent Variables

Recall that, in an equation, any variable whose value can be freely assigned is said to be an independent variable. Any variable whose value is determined once the other value or values have been assigned is said to be a dependent variable. If, in a linear equation, the independent variable is x and the dependent variable is y, and a solution to the equation is x=a and y=b, the solution is written as the

ORDERED PAIR: ((a, b))

Ordered Pair

In an ordered pair, ((a, b)), the first component, (a), gives the value of the independent variable, and the second component, (b), gives the value of the dependent variable.

We can use ordered pairs to show some solutions to the equation (y=6x−7).

Example (PageIndex{1})

((0, -7))

If (x = 0) and (y = -7), we get a true statement upon substitution and computation

(egin{array}{flushleft}
y&=&6x-7
-7&=&6(0)-7& ext{Is this correct?}
-7&=&-7& ext{Is this correct?}
-7&=&41& ext{Yes, this is correct}
end{array})

Example (PageIndex{2})

((8, 41))

If (x = 8) and (y = 41), we get a true statement upon substitution and computation

(egin{array}{flushleft}
y&=&6x-7
41&=&6(8)-7& ext{Is this correct?}
41&=&48-7& ext{Is this correct?}
41&=&41& ext{Yes, this is correct}
end{array})

Example (PageIndex{3})

((-4, -31))

If (x = 8) and (y = 41), we get a true statement upon substitution and computation

(egin{array}{flushleft}
y&=&6x-7
-31&=&6(-4)-7& ext{Is this correct?}
-31&=&-24-7& ext{Is this correct?}
-31&=&-31& ext{Yes, this is correct}
end{array})

These are only three of the infintely many solutions to this equation.

Sample Set A

Find a solution to each of the following linear equations in two variables and write the solution as an ordered pair.

Example (PageIndex{4})

(
y=3 x-6, ext { if } x=1
)
Substitute 1 for (x), compute, and solve for (y).
(
egin{aligned}
y &=3(1)-6
&=3-6 \
&=-3
end{aligned}
)
Hence, one solution is ((1,-3)).

Example (PageIndex{5})

(
y=15-4 x, ext { if } x=-10
)
Substitute -10 for (x), compute, and solve for (y).
(
egin{aligned}
y &=15-4(-10)
&=15+40 \
&=55
end{aligned}
)
Hence, one solution is ((-10,55)).

Example (PageIndex{6})

(
b=-9 a+21, ext { if } a=2
)
Substitute 2 for (a), compute, and solve for (b).
(
egin{aligned}
b &=-9(2)+21
&=-18+21 \
&=3
end{aligned}
)
Hence, one solution is ((2,3)).

Example (PageIndex{7})

(5 x-2 y=1, ext{ if } x=0)
Substitute 0 for (x), compute, and solve for (y).
(
egin{aligned}
5(0)-2 y &=1
0-2 y &=1
-2 y &=1
y &=-dfrac{1}{2}
end{aligned}
)
Hence, one solution is (left(0,-dfrac{1}{2} ight)).

Practice Set A

Find a solution to each of the following linear equations in two variables and write the solution as an ordered pair.

Practice Problem (PageIndex{1})

(y=7x−20), if (x=3)

Answer

((3, 1))

Practice Problem (PageIndex{2})

(m=−6n+1), if (n=2)

Answer

((2, −11))

Practice Problem (PageIndex{3})

(b=3a−7), if (a=0)

Answer

((0, −7))

Practice Problem (PageIndex{4})

(10x−5y−20=0), if (x=−8)

Answer

((−8, −20))

Practice Problem (PageIndex{5})

(3a+2b+6=0), if (a=−1)

Answer

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

Exercises

For the following problems, solve the linear equations in two variables.

Exercise (PageIndex{1})

(y=8x+14), if (x=1)

Answer

((1,22))

Exercise (PageIndex{2})

(y=−2x+1), if (x=0)

Exercise (PageIndex{3})

(y=5x+6), if (x=4)

Answer

((4,26))

Exercise (PageIndex{4})

(x+y=7), if (x=8)

Exercise (PageIndex{5})

(3x+4y=0), if (x=−3)

Answer

((-3, dfrac{9}{4}))

Exercise (PageIndex{6})

(−2x+y=1), if (x=dfrac{1}{2})

Exercise (PageIndex{7})

(5x−3y+1=0), if (x=−6)

Answer

((-6, -dfrac{29}{3}))

Exercise (PageIndex{8})

(−4x−4y=4), if (y=7)

Exercise (PageIndex{9})

(2x+6y=1), if (y=0)

Answer

((dfrac{1}{2}, 0))

Exercise (PageIndex{10})

(−x−y=0), if (y=dfrac{14}{3})

Exercise (PageIndex{11})

(y=x), if (x=1)

Answer

((1,1))

Exercise (PageIndex{12})

(x+y=0), if (x=0)

Exercise (PageIndex{13})

(y + dfrac{3}{4} = x), if (x = dfrac{9}{4})

Answer

((dfrac{9}{4}, dfrac{3}{2}))

Exercise (PageIndex{14})

(y+17=x), if (x=−12)

Exercise (PageIndex{15})

(−20y+14x=1), if (x=8)

Answer

((8, dfrac{111}{20}))

Exercise (PageIndex{16})

(dfrac{3}{5}y + dfrac{1}{4}x = dfrac{1}{2}), if (x = -3)

Exercise (PageIndex{17})

(dfrac{1}{5}x + y = -9), if (y = -1).

Answer

((−40,−1))

Exercise (PageIndex{18})

(y+7−x=0), if (x = *)

Exercise (PageIndex{19})

(2x+31y−3=0), if (x=a)

Answer

((a, dfrac{3-2a}{31}))

Exercise (PageIndex{20})

(436x+189y=881), if (x=−4231)

Exercise (PageIndex{21})

(y=6(x−7)), if (x=2)

Answer

((2,−30))

Exercise (PageIndex{22})

(y=2(4x+5)), if (x=−1)

Exercise (PageIndex{23})

(5y=9(x−3)), if (x=2)

Answer

((2, -dfrac{9}{5}))

Exercise (PageIndex{24})

(3y=4(4x+1)), if (x=−3)

Exercise (PageIndex{25})

(−2y=3(2x−5)), if (x=6)

Answer

((6, -dfrac{21}{2}))

Exercise (PageIndex{26})

(−8y=7(8x+2)), if (x=0)

Exercise (PageIndex{27})

(b=4a−12), if (a=−7)

Answer

((−7,−40))

Exercise (PageIndex{28})

(b=−5a+21), if (a=−9)

Exercise (PageIndex{29})

(4b−6=2a+1), if (a=0)

Answer

((0, dfrac{7}{4}))

Exercise (PageIndex{30})

(−5m+11=n+1), if (n=4)

Exercise (PageIndex{31})

(3(t+2)=4(s−9)), if (s=1)

Answer

((1, -dfrac{38}{3}))

Exercise (PageIndex{32})

(7(t−6)=10(2−s)), if (s=5)

Exercise (PageIndex{33})

(y=0x+5), if (x=1)

Answer

((1,5))

Exercise (PageIndex{34})

(2y=0x−11), if (x=−7)

Exercise (PageIndex{35})

(−y=0x+10), if (x=3)

Answer

((3,−10))

Exercise (PageIndex{36})

(−5y=0x−1), if (x=0)

Exercise (PageIndex{37})

(y=0(x−1)+6), if (x=1)

Answer

((1,6))

Exercise (PageIndex{38})

(y=0(3x+9)−1), if (x=12)

Exercise (PageIndex{39})

An examination of the winning speeds in the Indianapolis 500 automobile race from 1961 to 1970 produces the equation (y=1.93x+137.60), where (x) is the number of years from 1960 and (y) is the winning speed. Statistical methods were used to obtain the equation, and, for a given year, the equation gives only the approximate winning speed. Use the equation (y=1.93x+137.60) to find the approximate winning speed in

  1. 1965
  2. 1970
  3. 1986
  4. 1990
Answer

(a) Approximately 147 mph using ((5,147.25))
(b) Approximately 157 mph using ((10,156.9))
(c) Approximately 188 mph using ((26,187.78))
(d) Approximately 196 mph using ((30,195.5))

Exercise (PageIndex{40})

In electricity theory, Ohm’s law relates electrical current to voltage by the equation (y=0.00082x), where (x) is the voltage in volts and (y) is the current in amperes. This equation was found by statistical methods and for a given voltage yields only an approximate value for the current. Use the equation (y=0.00082x) to find the approximate current for a voltage of

  1. 6 volts
  2. 10 volts

Exercise (PageIndex{41})

Statistical methods have been used to obtain a relationship between the actual and reported number of German submarines sunk each month by the U.S. Navy in World War II. The equation expressing the approximate number of actual sinkings, (y), for a given number of reported sinkings, (x), is (y=1.04x+0.76). Find the approximate number of actual sinkings of German submarines if the reported number of sinkings is

  1. 4
  2. 9
  3. 10
Answer

(a) Approximately 5 sinkings using ((4,4.92))
(b) Approximately 10 sinkings using ((9,10.12))
(c) Approximately 11 sinkings using ((10,11.16))

Exercise (PageIndex{42})

Statistical methods have been used to obtain a relationship between the heart weight (in milligrams) and the body weight (in milligrams) of 10-month-old diabetic offspring of crossbred male mice. The equation expressing the approximate body weight for a given heart weight is (y=0.213x−4.44). Find the approximate body weight for a heart weight of

  1. 210 mg
  2. 245 mg

Exercise (PageIndex{43})

Statistical methods have been used to produce the equation (y=0.176x−0.64). This equation gives the approximate red blood cell count (in millions) of a dog’s blood, (y), for a given packed cell volume (in millimeters), (x). Find the approximate red blood cell count for a packed cell volume of

  1. 40 mm
  2. 42 mm
Answer

(a) Approximately 6.4 using ((40,6.4))
(b) Approximately 4.752 using ((42,7.752))

Exercise (PageIndex{44})

An industrial machine can run at different speeds. The machine also produces defective items, and the number of defective items it produces appears to be related to the speed at which the machine is running. Statistical methods found that the equation (y=0.73x−0.86) is able to give the approximate number of defective items, (y), for a given machine speed, (x). Use this equation to find the approximate number of defective items for a machine speed of

  1. 9
  2. 12

Exercise (PageIndex{45})

A computer company has found, using statistical techniques, that there is a relationship between the aptitude test scores of assembly line workers and their productivity. Using data accumulated over a period of time, the equation (y=0.89x−41.78) was derived. The (x) represents an aptitude test score and (y) the approximate corresponding number of items assembled per hour. Estimate the number of items produced by a worker with an aptitude score of

  1. 80
  2. 95
Answer

(a) Approximately 29 items using ((80,29.42))
(b) Approximately 43 items using ((95,42.77))

Exercise (PageIndex{46})

Chemists, making use of statistical techniques, have been able to express the approximate weight of potassium bromide, (W), that will dissolve in 100 grams of water at (T) degrees centigrade. The equation expressing this relationship is (W=0.52T+54.2). Use this equation to predict the potassium bromide weight that will dissolve in 100 grams of water that is heated to a temperature of

  1. 70 degrees centigrade
  2. 95 degrees centigrade

Exercise (PageIndex{47})

The marketing department at a large company has been able to express the relationship between the demand for a product and its price by using statistical techniques. The department found, by analyzing studies done in six different market areas, that the equation giving the approximate demand for a product (in thousands of units) for a particular price (in cents) is (y=−14.15x+257.11). Find the approximate number of units demanded when the price is

  1. $0.12
  2. $0.15
Answer

(a) Approximately 87 units using ((12,87.31))
(b) Approximately 45 units using ((15,44.86))

Exercise (PageIndex{48})

The management of a speed-reading program claims that the approximate speed gain (in words per minute), (G), is related to the number of weeks spent in its program, (W), is given by the equation (G=26.68W−7.44). Predict the approximate speed gain for a student who has spent

  1. 3 weeks in the program
  2. 10 weeks in the program

Exercises for Review

Exercise (PageIndex{49})

Find the product. ((4x−1)(3x+5)).

Answer

(12x^2+17x−5)

Exercise (PageIndex{50})

Find the product. ((5x+2)(5x−2))

Exercise (PageIndex{51})

Solve the equation (6[2(x−4)+1]=3[2(x−7)]).

Answer

(x=0)

Exercise (PageIndex{52})

Solve the inequality (−3a−(a−5)≥a+10).

Exercise (PageIndex{53})

Solve the compound inequality (−1<4y+11<27).

Answer

(−3


5.8: Linear Equations in Two Variables

Ann invested $12,000 in two bank accounts. One of the accounts pays 6% annual interest, and the other account pays 5% annual interest. If the combined interest earned in both accounts after a year was $700, how much money was invested in each account?

What are we trying to find in this problem?

We want to know the amount of money invested in each account-- in other words, we want to know the amount invested in the 6% account and the amount invested in the 5% account. Each of the things we are trying to find will be represented by a variable:

x = amount invested at 6%
y = amount invested at 5%

Since we have two variables to solve for, we will need to find a system of two equations to solve.

How do we find the two equations we need?

We are given two numbers in the problem:

$12,000 = total money invested in both accounts
$700 = total interest earned in both accounts

Let's start with the $12,000. Ann wants to split this money into two parts. We have chosen to call the two parts x and y. Since these two parts must total to $12,000, this gives us our first equation:

Now let's look at the $700, the interest earned on the two accounts together. Let's think about the formula for calculating simple interest :

Since the time period in this problem is one year, our simple interest equation becomes:

Interest = (Principle)(Rate)(1)
or
Interest = (Principle)(Rate)

Each account has a different amount of money invested in it (either x dollars or y dollars), and each account has a different interest rate (either 6% or 5%). This gives us the following:

Interest earned on x dollars = (x)(6%) = .06x

Interest earned on y dollars = (y)(5%) = .05y

The total interest earned in both accounts is $700, so our second equation is:

Interest earned on x dollars + interest earned on y dollars = total interest
.06x + .05y = 700

If we multiply both sides of this equation by 100 to clear the decimals, it becomes:
6x + 5y = 70,000

Now we'll solve the system of equations:

x + y = 12,000
6x + 5y = 70,000

Multiply the first equation by -5, then add the equations:

-5x - 5y = -60,000
6x + 5y = 70,000
x = 10,000

Ann invested $10,000 in the account that pays 6% interest.

To find the amount invested in the other account, substitute 10,000 for x in either of our equations. We'll choose the easier equation:

x + y = 12,000
10,000 + y = 12,000
y = 2,000

Ann invested $2,000 in the account that pays 5% interest.



5.8: Linear Equations in Two Variables

Thus far in this course, discussions of equations have been limited to linear equations in one variable. Linear equations which have two variables are common, and their solution involves extending some of the procedures which have already been introduced.

An outstanding characteristic of equations in two variables is their adaptability to graphical analysis. The rectangular coordinate system, which was introduced in chapter 3 of this course, is used in analyzing equations graphically. This system of vertical and horizontal lines, meeting each other at right angles and thus forming a rectangular grid, is often called the Cartesian coordinate system. It is named after the French philosopher and mathematician, Rene Descartes, who invented it.

The rectangular coordinate system is developed on a framework of reference similar to figure 3-2 in chapter 3 of this course. On a piece of graph paper, two lines are drawn intersecting each other at right angles, as in figure 12-1. The vertical line is usually labeled

with the capital letter Y and called the Y axis. The horizontal line is usually labeled with the capital letter X and called the X axis. The point where the X and Y axes intersect is called the ORIGIN and is labeled with the letter o. Above the origin, numbers measured along or parallel to the Y axis are positive below the origin they are negative. To the right of the origin, numbers measured along or parallel to the X axis are positive to the left they are negative.

A point anywhere on the graph may be located by two numbers, one showing the distance of the point from the Y axis, and the other showing the distance of the point from the X axis.

Figure 12-1 -Rectangular coordinate system. Point P (fig. 12-1) is 6 units to the right of the Y axis and 3 units above the X axis. We call the numbers that indicate the position of a point COORDINATES. The number indicating the distance of the point measured horizontally from the origin is the X coordinate (6 in this example), and the number indicating the distance of the point measured vertically from the origin (3 in this example) is the Y coordinate.

In describing the location of a point by means of rectangular coordinates, it is customary to place the coordinates within parentheses and separate them with a comma. The X coordinate is always written first. The coordinates of point P (fig. 12-1) are written (6, 3). The coordinates for point Q are (4, -5) for point R, they are (-5, -2) and for point S, they are (-8, 5).

Usually when we indicate a point on a graph, we write a letter and the coordinates of the point. Thus, in figure 12-1, for point S, we write S(-8, 5). The other points would ordinarily be written, P(6, 3), Q(4, -5), and R(-5, -2). The Y coordinate of a point is often called its ORDINATE and the X coordinate is often called its ABSCISSA.

The X and Y axes divide the graph into four parts called QUADRANTS. In figure 12-1, point P is in quadrant I, point S is in quadrant II, R is in quadrant III, and Q is in quadrant IV. In the first and fourth quadrants, the X coordinate is positive, because it is to the right of the origin. In the second and third quadrant it is negative, because it is to the left of the origin. Likewise, the Y coordinate is positive in the first and second quadrants, being above the origin it is negative in the third and fourth quadrants, being below the origin. Thus, we know in advance the signs of the coordinates of a point by knowing the quadrant in which the point appears. The signs of the coordinates in the four quadrants are shown in figure 12-1. Locating points with respect to axes is called PLOTTING. As shown with point P (fig. 12-l), plotting a point is equivalent to completing a rectangle that has segments of the axes as two of its sides with lines dropped perpendicularly to the axes forming the other two sides. This is the reason for the name "rectangular coordinates."

PLOTTING A LINEAR EQUATION

A linear equation in two variables may have many solutions. For example, in solving the equation 2x - y = 5, we can find an unlimited number of values of x for which there will be a corresponding value of y. When x is 4, y is 3, since (2 x 4) - 3 = 5. When x is 3, y is 1, and when x is 6, y is 7. When we graph an equation, these pairs of values are considered coordinates of points on the graph. The graph of an equation is nothing more than a line joining the points located by the various pairs of numbers that satisfy the equation.

To picture an equation, we first find several pairs of values that satisfy the equation. For example, for the equation 2x - y = 5, we assign several values to x and solve for y. A convenient way to find values is to first solve the equation for either variable, as follows:

Once this is accomplished, the value of y is readily apparent when values are substituted for x. The information derived may be recorded in a table such as table 12-1. We then lay off X and Y axes on graph paper, select some convenient unit distance for measurement along the axes, and then plot the pairs of values found for x and y as coordinates of points on the graph. Thus, we locate the pairs of values shown in table 12-1 on a graph, as shown in figure 12-2 (A).

Table 12-1.-Values of x and y in the equation

Figure 12-2.-Graph of 2x - y = 5.

Finally, we draw a line joining these points, as in figure 12-2 (B). It is seen that this is a straight line hence the name "linear equation." Once the graph is drawn, it is customary to write the equation it represents along the line, as shown in figure 12-2 (B).

It can be shown that the graph of an equation is the geometric representation of all the points whose coordinates satisfy the conditions of the equation. The line represents an infinite number of pairs of coordinates for this equation. For example, selecting at random the point on the line where x is 2: and y is 0 and substituting these values in the equation, we find that they satisfy it. Thus,

If two points that lie on a straight line can be located, the position of the line is known. The mathematical language for this is "Two points DETERMINE a straight line." We now know that the graph of a linear equation in two variables is a straight line. Since two points are sufficient to determine a straight line, a linear equation can be graphed by plotting two points and drawing a straight line through these points. Very often pairs of whole numbers which satisfy the equation can be found by inspection. Such points are easily plotted.

After the line is drawn through two points, it is well to plot a third point as a check. If this third point whose coordinates satisfy the equation lies on the line the graph is accurately drawn.


NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable| PDF Download

Solve the following equations.
(1) x – 2 = 7
(2) y + 3 = 10
(3) 6 = z + 2
(4) 3/7 + x = 17/7
(5) 6x = 12
(6) t/5 = 10
(7) 2x/3 = 18
(8) 1.6 = y/1.5
(9) 7x – 9 = 16
(10) 14y – 8 = 13
(11) 17 + 6p = 9
(12) x/3 + 1 = 7/15

(2) y + 3 = 10
⇒ y = 10 - 3
⇒ y = 7

(3) 6 = z + 2
⇒ z + 2 = 6
⇒ z = 6 - 2
⇒ z = 4

(4) 3/7 + x = 17/7
⇒ x = 17/7 - 3/7
⇒ x = (17 - 3)/7
⇒ x = 14/7
⇒ x = 2

(9) 7x – 9 = 16
⇒ 7x = 16 + 9
⇒ x = 25/7

(10) 14y – 8 = 13
⇒ 14y = 13 + 8
⇒ 14y = 21
⇒ y = 21/14
⇒ y = 3/2

1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?

Let the number be x.
A/q,
(x - 1/2) × 1/2 = 1/8
⇒ x/2 - 1/4 = 1/8
⇒ x/2 = 1/8 + 1/4
⇒ x/2 = 1/8 + 2/8
⇒ x/2 = (1+2)/8
⇒ x = 3/8 × 2
⇒ x = 6/8 = 3/4

2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?

Given,
Perimeter of rectangular swimming pool = 154 m
Let the breadth of rectangle be x.
A/q,
Length of the rectangle = 2x + 2
Perimeter = 2(length + breadth)
⇒ 2(2x + 2 + x) = 154 m
⇒ 2(3x + 2) = 154
⇒ 3x + 2 = 154/2
⇒ 3x = 77 - 2
⇒ x = 75/3
⇒ x = 25 m
Thus,
Breadth = x = 25 m
Length = 2x + 2 = 50 + 2 = 52 m

3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 62/15 cm. What is the length of either of the remaining equal sides?

Base of isosceles triangle = 4/3 cm
Perimeter of triangle = 62/15 cm
Let the length of equal sides of triangle be x.
A/q,
4/3 + x + x = 62/15 cm
⇒ 2x = (62/15 - 4/3) cm
⇒ 2x = (62 - 20)/15 cm
⇒ 2x = 42/15 cm
⇒ x = 42/15 × 1/2
⇒ x = 42/30 cm
⇒ x =7/5 cm
The length of either of the remaining equal sides are 7/5 cm.

4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Let the one of the number be x.
Then, other number will be x + 15
A/q,
x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 80
⇒ x = 40
First number = x = 40
Other number = x + 15 = 40 + 15 = 55

5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Let the two numbers be 5x and 3x.
A/q,
5x - 3x = 18
⇒ 2x = 18
⇒ x = 9
Thus, the numbers are 5x = 45 and 3x = 27.

6. Three consecutive integers add up to 51. What are these integers?

Let the three consecutive integers are x, x+1 and x+2.
A/q,
x + (x+1) + (x+2) = 51
⇒ 3x + 3 = 51
⇒ 3x = 51 - 3
⇒ 3x = 48
⇒ x = 16
Thus, the integers are x = 16, x+1 = 17 and x+2 = 18

7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Let the three consecutive multiples of 8 are 8x, 8(x+1) and 8(x+2).
A/q,
8x + 8(x+1) + 8(x+2) = 888
⇒ 8 (x + x+1 + x+2) = 888 (Taking 8 as common)
⇒ 8 (3x + 3) = 888
⇒ 3x + 3 = 888/8
⇒ 3x + 3 = 111
⇒ 3x = 108
⇒ x = 36
Thus, the three consecutive multiples of 8 are 8x = 8 × 36 = 288,
8(x+1) = 8 × (36+1) = 8 × 37 = 296 and
8(x+2) = 8 × (36+2) = 8 × 38 = 304

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

Let the three consecutive integers are x, x+1 and x+2.
A/q,
2x + 3(x+1) + 4(x+2) = 74
⇒ 2x + 3x +3 + 4x + 8 = 74
⇒ 9x + 11 = 74
⇒ 9x = 74 - 11
⇒ x = 63/9
⇒ x = 7
Thus, the numbers are x = 7, x+1 = 8 and x+2 = 9

9. The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Let the ages of Rahul and Haroon be 5x and 7x.
Four years later their age will be (5x + 4) and (7x + 4) respectively.
A/q,
(5x + 4) + (7x + 4) = 56
⇒ 5x + 4 + 7x + 4 = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 - 8
⇒ x = 48/12
⇒ x = 4
Present age of Rahul = 5x = 5ࡪ = 20
Present age of Haroon = 7x = 7ࡪ = 28

10. The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Let the number of boys be 7x and girls be 5x.
A/q,
7x = 5x + 8
⇒ 7x - 5x = 8
⇒ 2x = 8
⇒ x = 4
Number of boys = 7ࡪ = 28
Number of girls = 5ࡪ = 20

11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Let the age of Baichung’s father be x.
Therefor, Age of Baichung’s grandfather = (x+26)
and, Age of Baichung = (x-29)
A/q,
x + (x+26) + (x-29) = 135
⇒ 3x + 26 - 29 = 135
⇒ 3x = 135 + 3
⇒ 3x = 138
⇒ x = 138/3
⇒ x = 46
Age of Baichung’s father = x = 46
Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72
Age of Baichung = (x-29) = 46 - 29 = 17

12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Let the present age of Ravi be x.
Fifteen years later, Ravi age will be x+15 years.
A/q,
x + 15 = 4x
⇒ 4x - x = 15
⇒ 3x = 15
⇒ x = 5
Present age of Ravi = 5 years.

13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?

Let the rational be x.
A/q,
x × (5/2) + 2/3 = -7/12
⇒ 5x/2 + 2/3 = -7/12
⇒ 5x/2 = -7/12 - 2/3
⇒ 5x/2 = (-7 - 8)/12
⇒ 5x/2 = -15/12
⇒ 5x/2 = -5/4
⇒ x = (-5/4) × 2/5
⇒ x = -10/20
⇒ x = -1/2
Thus, the rational number is -1/2

14. Lakshmi is a cashier in a bank. She has currency notes of denominations �, 󌡶 and 󌡎, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is 𔜾,00,000. How many notes of each denomination does she have?

Let the numbers of notes of �, 󌡶 and 󌡎 be 2x, 3x and 5x respectively.
Value of � = 2x × 100 = 200x
Value of 󌡶 = 3x × 50 = 150x
Value of 󌡎 = 5x × 10 = 50x
A/q,
200x + 150x + 50x = 4,00,000
⇒ 400x = 4,00,000
⇒ x = 4,00,000/400
⇒ x = 1000
Numbers of � notes = 2x = 2000
Numbers of 󌡶 notes = 3x = 3000
Numbers of 󌡎 notes = 5x = 5000

15. I have a total of � in coins of denomination 𔜻, 𔜼 and 𔜿. The number of 𔜼 coins is 3 times the number of 𔜿 coins. The total number of coins is 160. How many coins of each denomination are with me?

Let the number of 𔜿 coins be x.
Therefor, number 𔜼 coins = 3x
and, number of 𔜻 coins = (160 - 4x)
Now,
Value of 𔜿 coins = x × 5 = 5x
Value of 𔜼 coins = 3x × 2 = 6x
Value of 𔜻 coins = (160 - 4x) × 1 = (160 - 4x)
A/q,
5x + 6x + (160 - 4x) = 300
⇒ 11x + 160 - 4x = 300
⇒ 7x = 140
⇒ x = 140/7
⇒ x = 20
Number of 𔜿 coins = x = 20
Number of 𔜼 coins = 3x = 60
Number of 𔜻 coins = (160 - 4x) = 160 - 80 = 80

16. The organisers of an essay competition decide that a winner in the competition gets a prize of � and a participant who does not win gets a prize of 󌡝. The total prize money distributed is 𔜽,000. Find the number of winners, if the total number of participants is 63.

Let the numbers of winner be x.
Therefor, number of participant didn't win = 63 - x
Total money given to the winner = x × 100 = 100x
Total money given to participant didn't win = 25×(63-x)
A/q,
100x + 25×(63-x) = 3,000
⇒ 100x + 1575 - 25x = 3,000
⇒ 75x = 3,000 - 1575
⇒ 75x = 1425
⇒ x = 1425/75
⇒ x = 19
Thus, the numbers of winners are 19.

Solve the following equations and check your results.
(1) 3x = 2x + 18
(2) 5t – 3 = 3t – 5
(3) 5x + 9 = 5 + 3x
(4) 4z + 3 = 6 + 2z
(5) 2x – 1 = 14 – x
(6) 8x + 4 = 3 (x – 1) + 7
(7) x = 4/5(x + 10)
(8) 2x/3 + 1 = 7x/15 + 3
(9) 2y + 5/3 = 26/3 - y
(10) 3m = 5 m – 8/5

(1) 3x = 2x + 18
⇒ 3x - 2x = 18
⇒ x = 18
Putting the value of x in RHS and LHS we get,
3 × 18 = (2 × 18)+18
⇒ 54 = 54
⇒ LHS = RHS

(2) 5t – 3 = 3t – 5
⇒ 5t - 3t = -5 + 3
⇒ 2t = -2
⇒ t = -1
Putting the value of t in RHS and LHS we get,
5×(-1) - 3 = 3×(-1) - 5
⇒ -5 - 3 = -3 - 5
⇒ -8 = -8
⇒ LHS = RHS

(3) 5x + 9 = 5 + 3x
⇒ 5x - 3x = 5 - 9
⇒ 2x = -4
⇒ x = -2
Putting the value of x in RHS and LHS we get,
5×(-2) + 9 = 5 + 3×(-2)
⇒ -10 + 9 = 5 + (-6)
⇒ -1 = -1
⇒ LHS = RHS

(4) 4z + 3 = 6 + 2z
⇒ 4z - 2z = 6 - 3
⇒ 2z = 3
⇒ z = 3/2
Putting the value of z in RHS and LHS we get,
(4 × 3/2) + 3 = 6 + (2 × 3/2)
⇒ 6 + 3 = 6 + 3
⇒ 9 = 9
⇒ LHS = RHS

(5) 2x – 1 = 14 – x
⇒ 2x + x = 14 + 1
⇒ 3x = 15
⇒ x = 5
Putting the value of x in RHS and LHS we get,
(2࡫) - 1 = 14 - 5
⇒ 10 - 1 = 9
⇒ 9 = 9
⇒ LHS = RHS

(6) 8x + 4 = 3 (x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
⇒ 8x - 3x = 4 - 4
⇒ 5x = 0
⇒ x = 0
Putting the value of x in RHS and LHS we get,
(8ࡦ) + 4 = 3 (0 – 1) + 7
⇒ 0 + 4 = 0 - 3 + 7
⇒ 4 = 4
⇒ LHS = RHS

(7) x = 4/5(x + 10)
⇒ x = 4x/5 + 40/5
⇒ x - 4x/5 = 8
⇒ (5x - 4x)/5 = 8
⇒ x = 8 × 5
⇒ x = 40
Putting the value of x in RHS and LHS we get,
40 = 4/5(40 + 10)
⇒ 40 = 4/5 × 50
⇒ 40 = 200/5
⇒ 40 = 40
⇒ LHS = RHS

(8) 2x/3 + 1 = 7x/15 + 3
2x/3 - 7x/15 = 3 - 1
⇒ (10x - 7x)/15 = 2
⇒ 3x = 2 × 15
⇒ 3x = 30
⇒ x = 10
Putting the value of x in RHS and LHS we get,
(2吆)/3 + 1 = (7吆)/15 + 3
⇒ 20/3 + 1 = 70/15 + 3
⇒ (20 + 3)/3 = (70 + 45)/15
⇒ 23/3 = 115/15
⇒ 23/3 = 23/3
⇒ LHS = RHS

(10) 3m = 5 m – 8/5
⇒ 3m - 5m = -8/5
⇒ -2m = -8/5

⇒-2m × 5 = -8
⇒ -10m = -8
⇒ m = -8/-10
⇒ m = 4/5
Putting the value of m in RHS and LHS we get,
3 × 4/5 = (5 × 4/5) – 8/5
⇒ 12/5 = 4 - 8/5
⇒ 12/5 = (20 - 8)/5
⇒ 12/5 = 12/5
⇒ LHS = RHS

1. Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Let the number be x.
A/q,
(x - 5/2) × 8 = 3x
⇒ 8x - 40/2 = 3x
⇒ 8x - 3x = 40/2
⇒ 5x = 20
⇒ x = 4
Thus, the number is 4.

2. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Let one of the positive number be x then other number will be 5x.
A/q,
5x + 21 = 2(x + 21)
⇒ 5x + 21 = 2x + 42
⇒ 5x - 2x = 42 - 21
⇒ 3x = 21
⇒ x = 7
One number = x = 7
Other number = 5x = 5࡭ = 35
The two numbers are 5 and 35.

3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Let the digit at tens place be x then digit at ones place will be (9-x).
Original two digit number = 10x + (9-x)
After interchanging the digits, the new number = 10(9-x) + x
A/q,
10x + (9-x) + 27 = 10(9-x) + x
⇒ 10x + 9 - x + 27 = 90 - 10x + x
⇒ 9x + 36 = 90 - 9x
⇒ 9x + 9x = 90 - 36
⇒ 18x = 54
⇒ x = 3
Original number = 10x + (9-x) = (10ࡩ) + (9-3) = 30 + 6 = 36
Thus, the number is 36.

4. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Let the digit at tens place be x then digit at ones place will be 3x.
Original two digit number = 10x + 3x
After interchanging the digits, the new number = 30x + x
A/q,
(30x + x) + (10x + 3x) = 88
⇒ 31x + 13x = 88
⇒ 44x = 88
⇒ x = 2
Original number = 10x + 3x = 13x = 13ࡨ = 26

5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Let the present age of Shobo be x then age of her mother will be 6x.
Shobo's age after 5 years = x + 5
A/q,
(x + 5) = 1/3 × 6x
⇒ x + 5 = 2x
⇒ 2x - x = 5
⇒ x = 5
Present age of Shobo = x = 5 years
Present age of Shobo's mother = 6x = 30 years

6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate � per metre it will cost the village panchayat � to fence the plot. What are the dimensions of the plot?

Let the length of the rectangular plot be 11x and breadth be 4x.
Rate of fencing per metre = �
Total cost of fencing = �
Perimeter of the plot = 2(l+b) = 2(11x + 4x) = 2吋x = 30x
Total amount of fencing = (30x × 100)
A/q,
(30x × 100) = 75000
⇒ 3000x = 75000
⇒ x = 75000/3000
⇒ x = 25
Length of the plot = 11x = 11吕 = 275
Breadth of the plot = 4x = 4吕 = 100

7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him 󌡶 per metre and trouser material that costs him 󌢞 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is 󌡨,600. How much trouser material did he buy?

Let 2x m of trouser material and 3x m of shirt material be bought by him.
Selling price of shirt material per metre = ₹ 50 + 50×(12/100) = ₹ 56
Selling price of trouser material per metre = ₹ 90 + 90×(10/100) = 󌢧
Total amount of sale = 󌡨,600
A/q,
(2x × 99) + (3x × 56) = 36600
⇒ 198x + 168x = 36600
⇒ 366x = 36600
⇒ x = 36600/366
⇒ x =100
Total trouser material he bought = 2x = 2𴠼 = 200 m.

8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Let the total number of deer be x.
Deer grazing in the field = x/2
Deer playing nearby = 3/4(x - x/2) = 3/4×x/2 = 3x/8
Deer drinking water = 9
A/q,
x/2 + 3x/8 + 9 = x
⇒ (4x + 3x)/8 + 9 = x
⇒ 7x/8 + 9 = x
⇒ x - 7x/8 = 9
⇒ (8x - 7x)/8 = 9
⇒ x = 9࡮
⇒ x = 72

9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Let the age of granddaughter be x and grandfather be 10x.
Also, he is 54 years older than her.
A/q,
10x = x + 54
⇒ 10x - x = 54
⇒ 9x = 54
⇒ x = 6
Age of grandfather = 10x = 10࡬ = 60 years.
Age of granddaughter = x = 6 years.

10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Let the age of Aman's son be x then age of Aman will be 3x.
A/q,
5(x - 10) = 3x - 10
⇒ 5x - 50 = 3x - 10
⇒ 5x - 3x = -10 + 50
⇒ 2x = 40
⇒ x = 20
Aman's son age = x = 20 years
Aman age = 3x = 3吐 = 60 years

Solve the following linear equations.
(1) x/2 - 1/5 = x/3 + 1/4
(2) n/2 - 3n/4 + 5n/6 = 21
(3) x + 7 - 8x/3 = 17/6 - 5x/2
(4) (x - 5)/3 = (x - 3)/5
(5) (3t - 2)/4 - (2t + 3)/3 = 2/3 - t
(6) m - (m - 1)/2 = 1 - (m - 2)/3

(1) x/2 - 1/5 = x/3 + 1/4
⇒ x/2 - x/3 = 1/4 + 1/5
⇒ (3x - 2x)/6 = (5 + 4)/20
⇒ 3x - 2x = 9/20 × 6
⇒ x = 54/20
⇒ x = 27/10

(2) n/2 - 3n/4 + 5n/6 = 21
⇒ (6n - 9n + 10n)/12 = 21
⇒ 7n/12 = 21
⇒ 7n = 21合
⇒ n = 252/7
⇒ n = 36

(3) x + 7 - 8x/3 = 17/6 - 5x/2
⇒ x - 8x/3 + 5x/2 = 17/6 - 7
⇒ (6x - 16x + 15x)/6 = (17 - 42)/6
⇒ 5x/6 = -25/6
⇒ 5x = -25
⇒ x = -5

(4) (x - 5)/3 = (x - 3)/5
⇒ x/3 - 15 = x/5 - 15
⇒ x/3 - x/5 = -15 + 15
⇒ (5x - 3x)/15 = 0
⇒ 2x/15 = 0
⇒ x = 0

(5) (3t - 2)/4 - (2t + 3)/3 = 2/3 - t
⇒ 3t/4 - 1/2 - (2t/3 + 1) = 2/3 - t
⇒ 3t/4 - 1/2 - 2t/3 - 1 = 2/3 - t
⇒ 3t/4 - 2t/3 + t = 2/3 + 1 + 1/2
⇒ (9t - 8t + 12t)/12 = 2/3 + 3/2
⇒ 13t/12 = (4 + 9)/6
⇒ 13t/12 = 13/6
⇒ t = 13/6 × 12/13
⇒ t = 12/6 = 2

(6) m - (m - 1)/2 = 1 - (m - 2)/3
⇒ m - (m/2 - 1/2) = 1 - (m/3 - 2/3)
⇒ m - m/2 + 1/2 = 1 - m/3 + 2/3
⇒ m - m/2 + m/3 = 1 + 2/3 - 1/2
⇒ m/2 + m/3 = 1/2 + 2/3
⇒ (3m + 2m)/6 = (3 + 4)/6
⇒ 5m/6 = 7/6
⇒ m = 7/6 × 6/5
⇒ m = 7/5

Simplify and solve the following linear equations.
(7) 3(t – 3) = 5(2t + 1)
(8) 15(y – 4) 𔃀(y – 9) + 5(y + 6) = 0
(9) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
(10) 0.25(4f – 3) = 0.05(10f – 9)

(7) 3(t – 3) = 5(2t + 1)
⇒ 3t - 9 = 10t + 5
⇒ 3t - 10t = 5 + 9
⇒ -7t = 14
⇒ t = 14/-7
⇒ t = -2

(8) 15(y – 4) 𔃀(y – 9) + 5(y + 6) = 0
⇒ 15y - 60 -2y + 18 + 5y + 30 = 0
⇒ 15y - 2y + 5y = 60 - 18 - 30
⇒ 18y = 12
⇒ y = 12/18
⇒ y = 2/3

(9) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z - 21 - 18z + 22 = 32z - 52 - 17
⇒ 15z - 18z - 32z = -52 - 17 + 21 - 22
⇒ -35z = -70
⇒ z = -70/-35
⇒ z = 2

(10) 0.25(4f – 3) = 0.05(10f – 9)
⇒ f - 0.75 = 0.5f - 0.45
⇒ f - 0.5f = -0.45 + 0.75
⇒ 0.5f = 0.30
⇒ f = 0.30/0.5
⇒ f = 30/5 = 6

Solve the following equations.
(1) (8x - 3)/3x = 2
(2) 9x/(7 - 6x) = 15
(3) z/(z + 15) = 4/9
(4) (3y + 4)/(2 - 6y) = -2/5
(5) (7y + 4)/(y + 2) = -4/3

(1) (8x - 3)/3x = 2
⇒ 8x/3x - 3/3x = 2
⇒ 8/3 - 1/x = 2
⇒ 8/3 - 2 = 1/x
⇒ (8 - 6)/3 = 1/x
⇒ 2/3 = 1/x
⇒ x = 3/2

(2) 9x/(7 - 6x) = 15
⇒ 9x = 15(7 - 6x)
⇒ 9x = 105 - 90x
⇒ 9x + 90x = 105
⇒ 99x = 105
⇒ x = 105/99 = 35/33

(3) z/(z + 15) = 4/9
⇒ z = 4/9(z + 15)
⇒ 9z = 4(z + 15)
⇒ 9z = 4z + 60
⇒ 9z - 4z = 60
⇒ 5z = 60
⇒ z = 12

(4) (3y + 4)/(2 - 6y) = -2/5
⇒ 3y + 4 = -2/5(2 - 6y)
⇒ 5(3y + 4) = -2(2 - 6y)
⇒ 15y + 20 = -4 + 12y
⇒ 15y - 12y = -4 - 20
⇒ 3y = -24
⇒ y = -8

(5) (7y + 4)/(y + 2) = -4/3
⇒ 7y + 4 = -4/3(y + 2)
⇒ 3(7y + 4) = -4(y + 2)
⇒ 21y + 12 = -4y - 8
⇒ 21y + 4y = -8 - 12
⇒ 25y = -20
⇒ y = 20/25 = 4/5

6. The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Let the age of Hari be 5x and Hari be 7x.
After 4years,
Age of Hari = 5x + 4
Age of Harry = 7x + 4
A/q,
(5x + 4)/(7x + 4) = 3/4
⇒ 4(5x + 4) = 3(7x + 4)
⇒ 20x + 16 = 21x + 12
⇒ 21x - 20x = 16 - 12
⇒ x = 4
Hari age = 5x = 5ࡪ = 20 years
Harry age = 7x = 7ࡪ = 28 years

7. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.

Let the numerator be x then denominator will be (x + 8).
A/q,
(x + 17)/(x + 8 - 1) = 3/2
⇒ (x + 17)/(x + 7) = 3/2
⇒ 2(x + 17) = 3(x + 7)
⇒ 2x + 34 = 3x + 21
⇒ 34 - 21 = 3x - 2x
⇒ 13 = x
The rational number is x/(x + 8) = 13/21


CBSE Class 9 Maths Extra Questions: Chapter 4 - Linear Equations In Two Variables (with Answers)

CBSE Class 9 Maths Extra Questions and Answers for Chapter 4 Linear Equations In Two Variables - Practice these important questions to obtain desired marks in exams in 2020-2021.

CBSE Class 9 Maths extra questions and answers for Chapter 4 - Linear Equations In Two Variables can are entirely based on the important concepts given in the NCERT book. Therefore these questions are best to assess your understanding of the concepts, giving you a chance of improvement in the weak areas. So, try to solve these questions seriously to prepare well for the exams and obtain desired marks.

CBSE Class 9 Maths Extra Questions for Chapter 4 - Linear Equations In Two Variables:

1. Find the equation of a line on which the point (1, 2) lies.

x + y = 3 is the equation on which the given point (1, 2) lies.

2. At which point the graph of linear equation x+2y = 2, cuts the y-axis?

Graph of the linear equation x+2y = 2, cuts the y-axis at point (0, 1).

3. Which of the following points lies on the line x = y:

4. x = 0 is the equation of the _______axis and y = 0 is the equation of the _______axis.

x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis.

5. Which of the following equations represents a line passing through the point (0, 0)?

6. The linear equation 3x − 5y = 14 has:

(a) Infinitely many solutions

(b) A unique solution

(c) Two solutions

(d) No solution

(a) Infinitely many solutions

7. The graph of x = 7 is a straight line parallel to ______ axis.

The graph of x = 7 is a straight line parallel to y - axis.

8. How does a solution of the linear equation change, both sides the equation are divided by a non-zero number?

Solution of the linear equation will remain the same.

9. Write the equation of a line which is parallel to y-axis and is at a distance of 5 units from the origin.

Required equation is: x + 5 = 0 or x − 5 = 0

10. Write the equations of two lines which lie in the same plane and are intersecting at the point (3, −7).

Equations of the required two lines are:

Students must go through the latest CBSE Syllabus for Class 9 Maths so that they can prepare according to the content prescribed by the board.


RD Sharma Solutions - Ex-3.10 Pair Of Linear Equations In Two Variables, Class 10, Maths Class 10 Notes | EduRev

Q1) Points A and B are 70km. apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7hrs, but if they travel towards each other, they meet in one hour. Find the speed of two cars.

Sol: We have to find the speed of car

Let x and y be two cars starting from points A and B respectively. Let the speed of car x be x km/hr and that of car y be y km/hr.

When two cars move in the same directions:

Suppose two cars meet at point Q, Then,

Distance travelled by car X = AQ

Distance travelled by car Y = BQ

It is given that two cars meet in 7 hours.

Therefore, Distance travelled by car X in 7 hours = 7x km

Distance traveled by car y in 7 hours = 7y km

7 x - 7 y = 70 Dividing both sides by common factor 7 we get,

When two cars move in opposite direction

Suppose two cars meet at point. Then,

Distance travelled by car x = AP ,

Distance travelled by car y = BP,

In this case, two cars meet in 1 hour

Therefore Distance travelled by car X in 1 hour = 1x km

Distance travelled by car Y in 1 hour = ly km

By solving equation (i) and (ii),

substituting x = 40 in equation (ii) we get

Hence, the speed of car starting from point A is 40km/hr.

The speed of car starting from point B is 30 km/hr.

2) A sailor goes 8km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and the speed of the current.

Sol: Let the speed of the sailor in still water be x km/hr and the speed of the current be y km/hr

Speed downstream = (x + y)km/hr

Now, Time taken to cover 8km downstream = hrs

Time taken to cover 8km upstream = hrs

But, time taken to cover 8 km downstream in 40 minutes or 40/60 hours that is 23hrs

= 2/3

Dividing both sides by common factor 2 we get

Time taken to cover 8km upstream in 1hour ,

= 1

By solving these equation (i)and (ii) we get,

Substitute x = 10 in equation (i) we get,

Hence, the speed of sailor is 10km/hr.

Speed of Current is 2km/hr.

3) The boat goes 30km upstream and 44km downstream in 10 hrs. In 13 hours it can go 40km upstream and 55km downstream. Determine the speed of stream and that of boat in still water.

Sol. Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr

Speed upstream = (x - y) km/hr

Speed downstream = (x + y) km/hr Now,

Time taken to cover 30 km upstream =

Time taken to cover 44 km downstream =

But total time of journey is 10 hours

+ = 10 ……..(i)

Time taken to cover 40 km upstream =

In this case total time of journey is given to be 13 hours

Therefore, = 13

Putting in equation (i) and (ii) we get


Solving Equations With Two Variables

This is part of a series of lessons for the quantitative reasoning section of the GRE revised General Test. In these lessons, we will learn:

  • Linear Equations in Two Variables
  • Solving Simultaneous Equations
  • Using the Substitution Method
  • Using the Elimination Method

Linear Equation In Two Variables

A linear equation in two variables, x and y, can be written in the form
ax + by = c
where x and y are real numbers and a and b are not both zero.

For example, 3x + 2y = 8 is a linear equation in two variables.

A solution of such an equation is an ordered pair of numbers (x, y) that makes the equation true when the values of x and y are substituted into the equation.

For example, both (2, 1) and (0, 4) are solutions of the equation but (2, 0) is not a solution. A linear equation in two variables has infinitely many solutions.

The following video shows how to complete ordered pairs to make a solution to linear equations.

Simultaneous Equations

If another linear equation in the same variables is given, it is usually possible to find a unique solution of both equations. Two equations with the same variables are called a system of equations , and the equations in the system are called simultaneous equations . To solve a system of two equations means to find an ordered pair of numbers that satisfies both equations in the system.

There are two basic methods for solving systems of linear equations, by substitution or by elimination.

Substitution Method

In the substitution method, one equation is manipulated to express one variable in terms of the other. Then the expression is substituted in the other equation.

For example, to solve the system of equations
3x + 2y = 2
y + 8 = 3x

Isolate the variable y in the equation y + 8 = 3x to get y = 3x – 8.

Then, substitute 3x – 8 for y into the equation 3x + 2y = 2.
3x + 2 (3x – 8) = 2
3x + 6x – 16 = 2
9x – 16 = 2
9x = 18

Substitute x = 2 into y = 3x – 8.to get the value for y
y = 3 (2) – 8
y = 6 – 8 = – 2

How to solve simultaneous equations using substitution?

Elimination Method

In the elimination method, the object is to make the coefficients of one variable the same in both equations so that one variable can be eliminated either by adding the equations together or by subtracting one from the other.

Consider the following example:
2x + 3y = –2 4x – 3y = 14

In this example the coefficients of y are already opposites (+3 and –3). Just add the two equations to eliminate y.

To get the value of y, we need to substitute x = 2 into the equation 2x + 3y = –2
2(2) + 3y = –2
4 + 3y = –2
3y = –6
y = –2

How to solve simultaneous equations using the substitution method and elimination (or combination) method

Example of the GRE Quantitative Comparison question that involves simultaneous equations

How to solve (linear) linear simultaneous equations by the method of elimination?
Four examples are given whereby the last example requires the multiplying of both of the equations before one of the variable can be eliminated.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

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Linear Equations in Two Variables



Examples, videos and solutions to help Grade 8 students learn how to solve linear equations with two variables.

New York State Common Core Math Grade 8, Module 4, Lesson 12

Lesson 12 Outcome

&bull Students use a table to find solutions to a given linear equation and plot the solutions on a coordinate plane.

Lesson 12 Summary

&bull A two-variable linear equation in the form ax + by = c is said to be in standard form.
&bull A solution to a linear equation in two variables is the ordered pair (x, y) that makes the given equation true. Solutions can be found by fixing a number for x and solving for y or fixing a number for y and solving for x.

Opening Exercise
Emily tells you that she scored 32 points in a basketball game with only two- and three-point baskets (no free throws).
How many of each type of basket did she score? Use the table below to organize your work.
Let x be the number of two-pointers and y be the number of three-pointers that Emily scored. Write an equation to represent the situation.

NYS Math Module 4 Grade 8 Lesson 12 Exercises

1. Find five solutions for the linear equation 2x = y - 4, and plot the solutions as points on a coordinate plane.

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Systems of Equations (Graphically)


Examples, solutions, videos and lessons to help Grade 8 students learn how to analyze and solve pairs of simultaneous linear equations.

A. Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.

B. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6.

C. Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair.

Suggested Learning Targets

  • I can identify the solution(s) to a system of two linear equations in two variables as the point(s) of intersection of their graphs.
  • I can describe the point(s) of intersection between two lines as the points that satisfy both equations simultaneously.
  • I can define "inspection."
  • I can solve a system of two equations (linear) in two unknowns algebraically.
  • I can identify cases in which a system of two equations in two unknowns has no solution.
  • I can identify cases in which a system of two equations in two unknowns has an infinite number of solutions.
  • I can solve simple cases of systems of two linear equations in two variables by inspection.
  • I can estimate the point(s) of intersection for a system of two equations in two unknowns by graphing the equations.
  • I can represent real-world and mathematical problems leading to two linear equations in two variables.

Solving Systems of Equations Graphically
Graphs intersect at one point.
The system is consistent and has one solution. Since neither equation is a multiple of the other, they are independent.

Graphs are parallel.
The system is inconsistent because there is no solution. Sine the equations are not equivalent, they are not independent.

Equations have the same graph.
The system is consistent and has infinite number of solutions. The equations are dependent since they are equivalent.

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