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

RD Sharma Class 9 Solutions Chapter 13 Ex 13.3 Free Download

RD Sharma Solutions Class 9 Chapter 13 Ex 13.3

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) Given here, x + y = 4

Thus we can write, y = 4 – x,

Now, putting x = 0 in y = 4 – x,

Therefore, y = 4

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

Thus, we can write the table presenting the abscissa and ordinates of points on the line represented as per given below;

X 0 4
Y 4 0

1

 

(ii) Given here, x – y = 2

So, y = x – 2

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

And putting x = 2 in y = x – 2, we get y = 0

Thus, we can write the table presenting the abscissa and ordinates of points on the line represented as per given below;

X 0 2
Y -2 0

2

(iii) Given here, – x + y = 6

So, y = 6 + x

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

So, y =6

Put x = -6 in y = 6+ x, to get y = 0

Thus, we can write the table presenting the abscissa and ordinates of points on the line represented as per given below;

X 0 -6
Y 6 0

3

(iv) Given to us here, y = 2x

Now, putting x = 1 in y = 2x

So, y = 2

Again putting, x = 3 in y = 2x

So, y = 6

Thus, we can write the table presenting the abscissa and ordinates of points on the line represented as per given below;

X 1 3
Y 2 6

4

(v) Given to us here, 3x + 5y = 15

So, 15 – 3x = 5y

Now, putting x = 0 in 5y = 15 – 3x,

So, 5y = 15

y =3

Putting x = 5 in 5y = 15 – 3x;

so, 5 y = 0

Thus, we can write the table presenting the abscissa and ordinates of points on the line represented as per given below;

X 0 5
Y 3 0

5

(vi) Given to us here;

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

\(\frac{3x – 2y}{6}=2\)

3x – 2y = 12

Thus, \(\frac{3x – 12}{2}=y\)

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

so,y = -6

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

so, y = 0

Thus, we can write the table presenting the abscissa and ordinates of points on the line represented as per given below;

X 0 4
Y -6 0

6

(vii) Given to us here,

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

So, x-2 = 3(Y-3)

x – 2 = 3y – 9

x = 3y – 7

Now, put x = 5 in x = 3y – 7, so that,

y = 4

puting x = 8 in x  = 3y – 7 , we have;

so y = 5

Thus, we can write the table presenting the abscissa and ordinates of points on the line represented as per given below;

X 5 8
Y 4 5

7

(viii) Given here, 2y = – x +1

so we can write, 1 – x = 2Y

Now, putting x =1 in1 – x = 2Y, we get;

y = 0

Again, putting x = 5 in 1 – x = 2Y , we get;

y = – 2

Thus, we can write the table presenting the abscissa and ordinates of points on the line represented as per given below;

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 can see that a = 3 and b = 12 is the solution of the following given equations;

4a – b = 0 and 3a – b + 3 = 0

Thus, we can get the equations of two lines passing through (3, 12) are, 4a – b = 0 and 3a – b + 3 = 0.

We know it already, infinite lines can be drawn, passing through the given point.

So, there are infinite 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 :

Let, total fare for covering the distance of a km is given by Rs.b

Then, as per the question;

b = 15 + 8(a – 1)

b = 15 + 8a – 8

b = 8a + 7

Therefore, b = 8a + 7 represents the linear equation for the given information.

 

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 :

Assume, total charges of Rs. 27, of which Rs. a for the first three days and Rs. b per day for 4 more days is given by

a + b ( 7 – 3 ) = 27

a + 4b = 27

Therefore, a + 4b = 27 represents the linear equation for the given information.

 

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  ‘ ba ‘,

Where ‘b’ represents the ten’s place.

And ‘a’ represents the unit’s place.

Now, the given number is in the form of 10b + a

Number produced by reversing the digits of the number is 10a + b

Given, the original number is 27 more than the number obtained by reversing its digits

So, 10b + a = 10a + b + 27

10b – b + a – 10a = 27

9b – 9a = 27

9 ( b – a ) = 27

b – a  = 27/9 = 3

a – b = 3, represents the required linear equation.

 

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 :

As per the question given, the number is in the form of ‘ ba’ ,

Here, ‘b’ represents the ten’s place and ‘a’ represents the units place

As per the statement given, the number is 10b + a.

On reversing the digits of the number, we get, 10x+ y

Given, the sum of the two numbers is 121.

So, ( 10b + a )+ ( l0a + b ) = 121

10b + b + a + 10a = 121

11b + 11a = 121

11 ( b + a ) = 121

a + b = 121/11 = 11

a + b = 11,  represents the required linear equation.

 

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

On plotting the points (3, 5) and (-1, 3), given to us, on a graph paper, we can draw line BC.

As you can see, Point A (1, 4) is already plotted on graph, by the intersecting lines.

Hence, 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: Given here, co-ordinates (1, – 1) and (-1, 1) as the solution of one of the equations.

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

10

(i) Given, y = x

Put x =1 and y = -1 ,

Thus, 1 ≠ -1

L.H.S  R.H.S

Putting x = -1  and y = 1 ,

we will get; -1 1

L.H.S ≠ R.H.S

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

(ii) Given,

x + y = 0

Putting x =1 and y = -1 , we get;

=> 1 + (-1) = 0

=> 0 = 0

L.H.S = R.H.S

Putting x = —1 and y = 1 ,we get

(-1)+ 1 = 0

0 = 0

L.H.S = R.H.S

Thus, the given solutions satisfy this equation.

 

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:

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

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

11

(i)  Given, y=x+2

Putting, x = – 1 and y = 3 ,we get

3 ≠ – 1 + 2

L.H.S ≠  R.H.S

Putting, x = 2 and y = 0 ,we get;

0 ≠ 4

L.H.S ≠  R.H.S

Thus, this solution does not satisfy the given equation.

(ii) Given, y = x – 2

Putting, x = —1 and y = 3 ,we get

3 ≠ – 1 – 2

L.H.S ≠  R.H.S

Putting, x = 2 and y = 0 ,we will get

0 = 0

L.H.S = R.H.S

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

(iii) Given, y = – x + 2

Putting, x = – 1 and y = 3,we get

3 = – ( – 1 ) + 2

L.H.S = R.H.S

Putting 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.

Hence, this is the equation whose graph has been given.

 

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

A10 :

Given, point (2,-2) lies on the given linear equation,

5x + ky = 4

We can see, the given point is the solution to the given equation.

Now, putting x = 2 and y = -2 in the given equation;

we get, 5x + ky = 4

5 × 2 + ( – 2 ) k = 4

2k = 10 – 4

2k = 6

k = 6/2

k = 3. Answer

 

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:

Given here,

2x +3y =12

Thus, \(y=\frac{12 – 2x}{3}\)

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

\(y=\frac{12 – 2\times 0}{3}\)

y = 4

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

\(y=\frac{12 – 2\times 6}{3}\)

y = 0

Thus, we have given table below exhibit the abscissa and ordinates of points on the line represented by the given equation;

X 0 6
Y 4 0

12

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

(i) Co-ordinates of the point whose y-coordinate 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) Given,

6x – 3y = 12

Thus,

y = (6x —12) /3

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

y = -4

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

y = 0

Thus, we have given table below exhibit 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 intersects the co-ordinate are y = – 4 at y axis and x = 2 at x axis.

(ii) Given here,

– x + 4y = 8

Thus,

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

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

y = 2

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

y = 0

Thus, we have given table below exhibit 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 intersects the co-ordinate are y = 2 at y-axis and x = —8 at x-axis.

(iii) Given here,

2x + y = 6

Thus,

y = 6 – 2x

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

y = 6

Putting x = 3 in y=6-2x, we get;

y=0

Thus, we have given table below exhibit 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 intersects the co-ordinate axes are y = 6 at y-axis and x =3 at x-axis.

(iv) Given here,

3x+2y+6=0

Thus,

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

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

y= – 3

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

y = 0

Thus, we have given table below exhibit 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 intersects 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 :

Given here;

2x + y = 6

Thus,

y = 6 – 2x

Now, putting x = 0 in y = 6 – 2x, we get;

y = 6

Putting, x =3 in y = 6— 2x, we get;

y = 0

Thus, we have given table below exhibit the abscissa and ordinates of points on the line represented by the given equation;

X 0 3
Y 6 0

17

The region surrounded by the graph, forms a triangle ABC.

Base of the triangle, AC = 6 units on y-axis

Height of triangle, BC = 3 units on x-axis

Therefore, Area of triangle ABC, A = 1/2 (Base x Height)

A = 1/2 ( AC x BC )

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:    

Given here,

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

4x +3y =12

Thus,

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

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

y =4

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

y = 0

Thus, we have given table below which shows 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.

Base of triangle, AC = 4 units on y axis.

Height of triangle, BC = 3 units on x axis.

Therefore,

Area of triangle ABC,  is given by

A = 1/2 (Base x Height)

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 given table below which shows 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 given table below which shows 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 given table below which shows 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 given table below which shows 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 given table below which shows 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 given table below which shows 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 given table below which shows 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:

The speed with which Aarushi was driving the car = 60 km/h. We will represent time here on X-axis and distance on Y-axis. in a graph:

25

Given, the car is travelling at a steady speed of 60 km/hr. This means the car drives 60 km distance per hour. Thus, we get a straight line graph here.

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

Joining points 0 and M and extending the line in both directions.

Let us 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 the y-axis at 30. Thus, in 12hr, distance travelled by 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

 

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