RD Sharma Solutions Class 9 Linear Equation In Two Variables Exercise 13.3

RD Sharma Solutions Class 9 Chapter 13 Exercise 13.3

RD Sharma Class 9 Solutions Chapter 13 Ex 13.3 Download

Q 1: Draw the graph of each of the following linear equations in two variables:

(i) x + y = 4          (ii) x – y = 2          (iii) -x + y = 6      (iv) y = 2x(v) 3x + 5y = 15

(vi) \(\frac{x}{2}-\frac{y}{3}=2\)         (vii) \(\frac{x-2}{3}\) = y- 3

(viii) 2y = -x +1

A 1 :

(i) We are given, x + y = 4

We get, y = 4 – x,

Now, substituting x = 0 in y = 4 – x,

we get y = 4

Substituting x = 4 in y = 4 — x, we get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given table

X 0 4
Y 4 0

1

 

(ii) We are given, x – y = 2

We get, y = x – 2

Now, substituting x = 0 in y= x – 2, we get y = – 2

Substituting x = 2 in y = x – 2, we get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 2
Y -2 0

2

(iii) We are given, – x + y = 6

We get, y = 6 + x

Now, substituting x = 0 in y = 6 + x,

We get y =6

Substituting x = -6 in y = 6+ x, we get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation.

X 0 -6
Y 6 0

3

(iv) We are given, y = 2x

Now, substituting x = 1 in y = 2x

We get y = 2

Substituting x = 3 in y = 2x

We get y = 6

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 1 3
Y 2 6

4

(v) We are given, 3x + 5y = 15

We get, 15 – 3x = 5y

Now, substituting x = 0 in 5y = 15 – 3x,

We get; 5y = 15

y =3

Substituting x = 5 in 5y = 15 – 3x

we get 5 y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 5
Y 3 0

5

(vi) we are given.

\(\frac{x}{2}-\frac{y}{3}=2\)

\(\frac{3x – 2y}{6}=2\)

3x – 2y = 12

We get, \(\frac{3x – 12}{2}=y\)

Now, substituting x = 0 in \(\frac{3x – 12}{2}=y\)

We get y = -6

Substituting x = 4 in \(\frac{3x – 12}{2}=y\)

We get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 4
Y -6 0

6

(vii) We are given,

\(\frac{x-2}{3}=y-3\)

We get, x-2 = 3(Y-3)

x – 2 = 3y – 9

x = 3y – 7

Now, substituting x = 5 in x = 3y – 7,

We get; y = 4

Substituting x = 8 in x  = 3y – 7 ,

We get; y = 5

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 5 8
Y 4 5

7

(viii) We are given, 2y = – x +1

We get, 1 – x = 2Y

Now, substituting x =1 in1 – x = 2Y, we get

y = 0

Substituting x = 5 in 1 – x = 2Y , we get

y = – 2

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 1 5
Y 0 -2

8

 

Q 2: Give the equations of two lines passing through ( 3, 12). How many more such lines are there, and why?

A 2:

We observe that x = 3 and y = 12 is the solution of the following equations

4x – y = 0 and 3x – y + 3 = 0

So, we get the equations of two lines passing through (3, 12) are, 4x – y = 0 and 3x – y + 3 = 0.

We know that passing through the given point infinitely many lines can be drawn.

So, there are infinitely many lines passing through (3, 12)

 

Q 3 : A three-wheeler scooter charges Rs 15 for first kilometer and Rs 8 each for every subsequent kilometer. For a distance of x km, an amount of Rs y is paid. Write the linear equation representing the above information.

A 3 :

Total fare of Rs y for covering the distance of x km is given by

y = 15 + 8(x – 1)

y = 15 + 8x – 8

y = 8x + 7

Where, Rs y is the total fare (x – 1) is taken as the cost of first kilometer is already given

Rs 15 and 1 has to subtracted from the total distance travelled to deduct the cost of first

Kilometer.

 

Q 4 : A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Aarushi paid Rs 27 for a book kept for seven days. If fixed charges are Rs x and per day charges are Rs y. Write the linear equation representing the above information.

A  4 :

Total charges of Rs 27 of which Rs x for first three days and Rs y per day for 4 more days is given by

x + y ( 7 – 3 ) = 27

x + 4y = 27

Here, (7 —3) is taken as the charges for the first three days are already given at Rs x and we have to find the charges for the remaining four days as the book is kept for the total of 7 days.

 

Q5: A number is 27 more than the number obtained by reversing its digits. lf its unit’s and ten’s digit are x and y respectively, write the linear equation representing the statement.

A5:

The number given to us is in the form of  ‘ yx ‘,

Where y represents the ten’s place of the number

And x represents the unit’s place of the number.

Now, the given number is 10y + x

Number obtained by reversing the digits of the number is 10x + y

It is given to us that the original number is 27 more than the number obtained by reversing its digits

So, 10y + x = 10x + y + 27

10y – y + x – 10x = 27

9y – 9x = 27

9 ( y – x ) = 27

y – x  = 27/9 = 3

x – y + 3 = 0

 

Q6: The Sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and tens digit of the number are x and y respectively, then write the linear equation representing the above statement.

A6 :

The number given to us is in the form of ‘ yx’ ,

Where y represents the ten’s place of the number and x represents the units place of the number

Now, the given number is 10y + x

Number obtained by reversing the digits of the number is 10x+ y

It is given to us that the sum of these two numbers is 121

So, ( 10y + x )+ ( l0x + y ) = 121

10y + y + x + 10x = 121

11y + 11x = 121

11 ( y + x ) = 121

x + y = 121/11 = 11

x + y = 11

 

Q7 : Plot the Points (3,5) and (-1,3) on a graph paper and verify that the straight line passing through the points, also passes through the point (1,4)

A7:

9

By plotting the given points (3, 5) and (-1, 3) on a graph paper, we get the line BC.

We have already plotted the point A (1, 4) on the given plane by the intersecting lines.

Therefore, it is proved that the straight line passing through (3, 5) and (-1, 3) also passes through A (1, 4).

 

Q8: From the choices given below, choose the equations whose graph is given in fig

(i) y = x (ii) x + y = 0 (iii) y = 2x (iv) 2 + 3y = 7x

 Ans: We are given co-ordinates (1, – 1) and (-1, 1) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

10

(i) We are given, y = x

Substituting x =I and y = -1 ,

we get; 1 \(\neq\) -1

L.H.S \(\neq\) R.H.S

Substituting x = -1  and y = 1 ,

we get; -1 \(\neq\)1

L.H.S \(\neq\)R.H.S

Therefore, the given equation y = x does not represent the graph in the figure.

(ii) We are given,

x + y = 0

Substituting x =1 and y = -1 , we get

=> 1 + (-1) = 0

=> 0 = 0

L.H.S = R.H.S

Substituting x = —1 and y = 1 ,we get

(-1)+ 1 = 0

0 = 0

L.H.S = R.H.S

Therefore, the given solutions satisfy this equation.

Thus, it is the equation whose graph is given.

 

Q9: From the choices given below, choose the equation whose graph is given fig:

(i) y = x + 2 (ii) y = x – 2 (iii)y = – x + 2 (iv) x + 2y = 6

A9:

We are given co-ordinates (-1, 3) and (2, 0) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

11

(i)            We are given, y=x+2

Substituting x = – 1 and y = 3 ,we get

3 \(\neq\) – 1 + 2

L.H.S \(\neq\) R.H.S

Substituting x = 2 and y = 0 ,we get

0 \(\neq\) 4

L.H.S \(\neq\)R.H.S

Therefore, the given solution does not satisfy this equation.

(ii)           We are given, y = x – 2

Substituting x = —1 and y = 3 ,we get

3 = – 1 – 2

L.H.S \(\neq\) R.H.S

Substituting x = 2 and y = 0 ,we get

0 = 0

L.H.S = R.H.S

Therefore, the given solutions does not completely satisfy this equation.

(iii) We are given, y = – x + 2

Substituting x = – 1 and y = 3,we get

3 = – ( – 1 ) + 2

L.H.S = R.H.S

Substituting x = 2 and y = 0 ,we get

0 = -2 + 2

0 = 0

L.H.S = R.H.S

Therefore, the given solutions satisfy this equation.

Thus, it is the equation whose graph is given.

 

Q 10 : If the point (2, -2) lies on the graph of linear equation, 5x + 4y = 4, find the value of k.

A10 :

It is given that the point (2,-2) lies on the given equation,

5x + ky = 4

Clearly, the given point is the solution of the given equation.

Now, Substituting x = 2 and y = – 2 in the given equation, we get 5x + ky = 4

5 x 2 + ( – 2 ) k = 4

2k = 10 – 4

2k = 6

k = 6/2

k = 3

 

Q 11 : Draw the graph of equation 2x + 3y = 12. From the graph, find the co ordinates of the point:

(i) whose y-coordinate is 3          (ii)whose x coordinate is -3

A11:

We are given,

2x +3y =12

We get, \(y=\frac{12 – 2x}{3}\)

Substituting, x = 0 in \(y=\frac{12 – 2x}{3}\), we get

\(y=\frac{12 – 2\times 0}{3}\)

y = 4

Substituting x = 6 in \(y=\frac{12 – 2x}{3}\)

\(y=\frac{12 – 2\times 6}{3}\)

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 6
Y 4 0

12

By plotting the given equation on the graph, we get the point B (0, 4) and C (6,0).

(i) Co-ordinates of the point whose y axis is 3 are A (3/2, 3)

(ii) Co-ordinates of the point whose x -coordinate is —3 are D (-3, 6)

 

Q 12: Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:

 (i) 6x – 3y = 12 (ii) – x + 4y = 8 (iii) 2x + y = 6 (iv) 3x + 2y + 6 = 0

A12 :

(i) We are given,

6x – 3y = 12 We get,

y = (6x —12) /3

Now, substituting x = 0 in y = – (6x – 12)/3 we get

y =- 4

Substituting x = 2 in y = (- 6x —12)/3, we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x 0 2
y -4 0

13

Co-ordinates of the points where graph cuts the co-ordinate axes are y = – 4 at y axis and x = 2 at x axis. (ii) We are given,

– x + 4y = 8

We get,

y = \(\frac{8 + x}{4}\)

Now, substituting x = 0 in y = \(\frac{8 + x}{4}\), we get

y = 2

Substituting x = -8 in y = \(\frac{8 + x}{4}\) ,we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 -8
Y 2 0

14

Co-ordinates of the points where graph cuts the co-ordinate axes are y = 2 at y axis and x = —8 at x axis.

(iii) We are given,

2x + y = 6

We get, y = 6 – 2x

Now, substituting x = 0 in y = 6 -2x we get

y = 6

Substituting x = 3 in y=6-2x, we get

y=0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 3
Y 6 0

15

Co-ordinates of the points where graph cuts the co-ordinate axes are y = 6 at y axis and x =3 at x axis.

(iv) We are given,

3x+2y+6=0

We get, \(y=\frac{-(6+3x)}{2}\)

Now, substituting x = 0 in \(y=\frac{-(6+3x)}{2}\),we get

y= – 3

Substituting x = —2 in \(y=\frac{-(6+3x)}{2}\), we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 -2
y -3 0

 

16

Co-ordinates of the points where graph cuts the co-ordinate axes are y = – 3 at y axis and x = – 2 at x axis.

 

Q 13 : Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.

A13 :

We are given,

2x + y = 6

We get,

y = 6 – 2x

Now, substituting x = 0 in y = 6 – 2x,

we get y = 6

Substituting x =3 in y = 6— 2x,

we get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 3
Y 6 0

17

The region bounded by the graph is ABC which forms a triangle.

AC at y axis is the base of triangle having AC = 6 units on y axis.

BC at x axis is the height of triangle having BC = 3 units on x axis.

Therefore, Area of triangle ABC, say A is given by A = (Base x Height)/2

A = ( AC x BC )/2

A = ( 6 x 3 )/2

A = 9 sq. units

 

Q 14 : Draw the graph of the equation \(\frac{x}{3}+\frac{y}{4}=1\) . Also, find the area of the triangle formed by 3 4 the line and the coordinates axes.

A14:      

We are given.

\(\frac{x}{3}+\frac{y}{4}=1\)

4x +3y =12

We get,

\(y=\frac{12-4x}{3}\)

Now, substituting x = 0 in \(y=\frac{12-4x}{3}\),we get

y =4

Substituting x=3 in \(y=\frac{12-4x}{3}\), we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 3
Y 4 0

18

The region bounded by the graph is ABC which forms a triangle.

AC at y axis is the base of triangle having AC = 4 units on y axis.

BC at x axis is the height of triangle having BC = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

A = (Base x Height)/2

A= (AC x BC)/2

A = ( 4 x 3 )/2

A = 6 sq. units

 

Q 15 : Draw the graph of y = | x |.

A15:

We are given,

y = |x|

Substituting x = 1 , we get y = 1

Substituting x = -1, we get y = 1

Substituting x = 2, we get y = 2

Substituting x = -2, we get y = 2

19

For every value of x, whether positive or negative, we get y as a positive number.

 

Q 16:  Draw the graph of  y = |x| + 2.

A16:

We are given,

Y = |x|+2

Substituting x = 0 we get y=2

Substituting x = I , we get y=3

Substituting x =-1 , we get Y = 3

Substituting x = 2, we get y = 4

Substituting x = -2, we get y = 4

20

For every value of x, whether positive or negative, we get y as a positive number and the minimum value of y is equal to 2 units.

 

Q 17 : Draw the graphs of the following linear equations on the same graph paper: 2x + 3y = 12, x – y = 1 Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis. Also, find the area of the triangle.

A17:

We are given,

2x +3y =12

We get, \(y=\frac{12-2x}{3}\)

Now, substituting x = 0 in \(y=\frac{12-2x}{3}\), we get

y =4

Substituting x=6 in \(y=\frac{12-2x}{3}\), we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 6
Y 4 0

 

Plotting A(0,4) and E(6,0) on the graph and by joining the points , we obtain the graph of equation

2x+3y=12.

We are given,

x – y = 1

We get, y = x – 1

Now, substituting x = 0 in y=x-1,

we get y = -1

Substituting x in y=x-1,

we get y = -2

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 -1
Y -1 -2

 

Plotting D(0, ) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation          x— y = I .

21

By the intersection of lines formed by 2x + 3y = 12 and x—y=1 on the graph, triangle ABC is formed on y axis.

Therefore, AC at y axis is the base of triangle ABC having AC = 5 units on y axis.

Draw FE perpendicular from F on y axis. FE parallel to x axis is the height of triangle ABC having FE = 3 units on x axis.

Therefore, Area of triangle ABC, say A is given by A = (Base x Height)/2 =(AC x FE)/2 = (5×3)/2    =>15/2 = 7.5 sq. units

 

Q 18 : Draw the graphs of the linear equations 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0. Find the area bounded by these lines and x-axis.

A18 :

We are given, 4x – 3y + 4 = 0

We get, \(y=\frac{4x+4}{3}\)

Now, substituting x = 0 in \(y=\frac{4x+4}{3}\), we get

Substituting x = -I in \(y=\frac{4x+4}{3}\)

we get y =0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x 0 -1
y 4/3 0

 

Plotting E(0, 4/3 ) and A (-1, 0) on the graph and by joining the points, we obtain the graph of equation

4x – 3y + 4 = 0.

We are given, 4x + 3y – 20 = 0

We get,

\(y=\frac{20-4x}{3}\)

Now, substituting x = 0 in \(y=\frac{20-4x}{3}\), we get

y = 7

Substituting x = 5 in \(y=\frac{20-4x}{3}\), we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 5
Y 20/3 0

 

Plotting D (0,20/3 ) and B(5,0) on the graph and by joining the points , we obtain the graph of equation 4x +3y – 20 = 0.

22

By the intersection of lines formed by 4x-3y + 4 = 0 and 4x+ 3y – 20 = 0 on the graph,

Triangle ABC is formed on x axis. Therefore, AB at x axis is the base of triangle ABC having AB = 6 units on x axis.

Draw CF perpendicular from C on x axis.

CF parallel to y axis is the height of triangle ABC having CF = 4 units on y axis.

Therefore, Area of triangle ABC, say A is given by

A = (Base x Height)/2

A =( AB x CF)/2

A = ( 6 x 4 )/2

k= 12 sq. units

 

Q19  :  The path of a train A is given by the equation 3x + 4y -12 = 0 and the path of another train B is given by the equation 6x + 8y – 48 = 0. Represent this situation graphically.

A19:

We are given the path of train A, 3x + 4y – 12 = 0

We get,

\(y=\frac{12-3x}{4}\)

Now, substituting x = 0 in \(y=\frac{12-3x}{4}\), we get

Y=3

Substituting x = 4 in \(y=\frac{12-3x}{4}\), we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 4
Y 3 0

 

Plotting A(4,0) and E(0,3) on the graph and by joining the points , we obtain the graph of equation 3x+4y-12 = 0.

We are given the path of train B,

6x + 8y – 48 = 0

We get, \(y=\frac{48-6x}{8}\)

Now, substituting x = 0 in \(y=\frac{48-6x}{8}\) ,we get

y=6

Substituting x = 8 in \(y=\frac{48-6x}{8}\), we get

y=0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 

X 0 8
Y 6 0

Plotting C(0,6) and D(8,0) on the graph and by joining the points , we obtain the graph of equation 6x+8y-48=0

23

 

Q 20 : Ravish tells his daughter Aarushi, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.

A20:

We are given the present age of Ravish as y years and Aarushi as x years.

Age of Ravish seven years ago = y – 7

Age of Aarushi seven years ago = x – 7

It has already been said by Ravish that seven years ago he was seven times old then Aarushi was then

So, y – 7 = 7 ( x – 7 )

Y – 7 = 7 x – 49

7x  – y = – 7 + 49

7x – y – 42 = 0                                    ——(1)

Age of Ravish three years from now = y + 3

Age of Aarushi three years from now = x+3

It has already been said by Ravish that three years from now he will be three times older then Aarushi will be then So,

Y + 3 = 3 ( x + 3 )

y + 3 = 3x + 9

3x + 9 – y – 3 = 0

3x – y + 6 = 0                                      —–(2)

(1) and (2) are the algebraic representation of the given statement.

We are given,

7x – y- 42 = 0

We get,

Y = 7x – 42

Now, substituting x = 0 in y = 7x —42,

we get y = -42

Substituting x = 6 in y = 7x – 42,

we get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 6
Y -42 0

We are given,

3x – y + 6 =0

We get,

Y = 3x + 6

Now, substituting x=0 in y = 3x + 6,

We get y = 6

Substituting x= —2 in y = 3x + 6,

We get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X 0 -2
Y 6 0

24

The red -line represents the equation 7x—y-42 =0.

The blue-line represents the equation 3x —y +6 =0.

 

Q21:  Aarushi was driving a car with uniform speed of 60 km/h. Draw distance-time graph From the graph, find the distance travelled by Aarushi in  (i) \(2\frac{1}{2}\) Hours (ii) \(\frac{1}{2}\) Hour

A21:

Aarushi is driving the car with the uniform speed of 60 km/h. We represent time on X-axis and distance on Y-axis Now, graphically

25

We are given that the car is travelling with a uniform speed 60 km/hr. This means car travels 60 km distance each hour. Thus the graph we get is of a straight line.

Also, we know when the car is at rest, the distance travelled is 0 km, speed is 0 km/hr and the time is also 0 hr. Thus, the given straight line will pass through O ( 0 , 0 ) and M ( 1 , 60 ).

Join the points 0 and M and extend the line in both directions.

Now, we draw a dotted line parallel to y-axis from x = 12 that meets the straight line graph at L from which we draw a line parallel to x-axis that crosses y-axis at 30. Thus, in 12hr, distance travelled by the car is 30 km.

Now, we draw a dotted line parallel to y-axis from x = 212 that meets the straight line graph at N from which we draw a line parallel to x-axis that crosses y-axis at 150. Thus, in 212hr, distance travelled by the car is 150 km.

(i) Distance = Speed x Time Distance travelled in \(2\frac{1}{2}\) hours is given by

Distance=60 x \(2\frac{1}{2}\)

Distance = 60 x 5/2

Distance = 150 Km

(ii) Distance = Speed x Time Distance travelled in \(\frac{1}{2}\) hour is given by

Distance=60 x \(2\frac{1}{2}\)

Distance = 30 km