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

*(vi)**\(\frac{x}{2}-\frac{y}{3}=2\)Â Â Â Â Â Â Â Â (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 â€” 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 |

(ii) Given here, x – y = 2

So, y = x â€“ 2

Now, putting x = 0 in y= x â€“ 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 |

(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 |

(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

X | 1 | 3 |

Y | 2 | 6 |

(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

X | 0 | 5 |

Y | 3 | 0 |

(vi) Given to us here;

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

\(\frac{3x – 2y}{6}=2\)

3x – 2y = 12

Thus,Â \(\frac{3x – 12}{2}=y\)

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

X | 0 | 4 |

Y | -6 | 0 |

(vii) Given to us here,

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

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

x â€“ 2 = 3y â€“ 9

x = 3y – 7

Now, put x = 5 in x = 3y â€“ 7, so that,

y = 4

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

so y = 5

X | 5 | 8 |

Y | 4 | 5 |

(viii) Given here, 2y = – x +1

so we can write, 1 â€“ x = 2Y

Now, putting x =1 in1 â€“ x = 2Y, we get;

y = 0

Again, putting x = 5 in 1 â€“ x = 2Y , we get;

y = – 2

X | 1 | 5 |

Y | 0 | -2 |

*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 â€“ b = 0 and 3a â€“ 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, bÂ = 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.*

*AÂ 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 â€“ 3 ) = 27

a + 4b = 27

Therefore,Â 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 Â â€˜ ba â€˜,

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 â€“ b + a – 10a = 27

9b – 9a = 27

9 ( b â€“ a ) = 27

b â€“ aÂ = 27/9 = 3

a â€“ 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.*

** A6Â **:

As per the question given, the number is in the form of â€˜ baâ€™ ,

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,Â 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:**

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 **

*Â ***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.

(i) Given, y = x

Put x =1 and y = -1 ,

Thus, 1 â‰ -1

L.H.S â‰ Â R.H.S

Putting x = -1 Â and y = 1 ,

we will get; -1 â‰ 1

L.H.S â‰ 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 = â€”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 â€“ 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.

(i)Â Â Given, y=x+2

Putting, x = – 1 and y = 3 ,we get

3 â‰ – 1 + 2

L.H.S â‰ Â R.H.S

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

0 â‰ 4

L.H.S â‰ Â R.H.S

Thus, this solution does not satisfy the given equation.

(ii) Given, y = x â€“ 2

Putting, x = â€”1 and y = 3 ,we get

3Â â‰ – 1 â€“ 2

L.H.S â‰ Â 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.**

**A10Â **:

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

5Â Ã— 2 + ( – 2 ) k = 4

2k = 10 â€“ 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 Â Â Â Â Â Â Â Â (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 |

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 â€”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:**

**Â (i) 6x â€“ 3y = 12 (ii) – x + 4y = 8 (iii) 2x + y = 6 (iv) 3x + 2y + 6 = 0**

**A12 : **

(i) Given,

6x – 3y = 12

Thus,

y = (6x â€”12) /3

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

y = -4

Putting, x = 2 in y = (- 6x â€”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 |

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 |

Co-ordinates of the points where graph intersects the co-ordinate are y = 2 at y-axis and x = â€”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

X | 0 | 3 |

Y | 6 | 0 |

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 = â€”2 in \(y=\frac{-(6+3x)}{2}\), we get;

y = 0

X | 0 | -2 |

y | -3 | 0 |

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 = 6â€” 2x, we get;

y = 0

X | 0 | 3 |

Y | 6 | 0 |

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:Â Â Â Â **

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 |

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,Â 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

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

*Q 16:Â Draw the graph of Â 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

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 â€“ y = 1

We get, y = x â€“ 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 Â Â Â Â Â Â Â Â xâ€” y = I .

By the intersection of lines formed by 2x + 3y = 12 and xâ€”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 Â Â =>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 â€“ 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

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.

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

*Â *

*Q19 Â : Â 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

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 â€“ 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

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

*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 â€“ 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 â€“ 7 )

Y â€“ 7 = 7 x â€“ 49

7xÂ – y = – 7 + 49

7x â€“ y â€“ 42 = 0 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ——(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 â€“ y â€“ 3 = 0

3x â€“ y + 6 = 0 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â —–(2)

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

We are given,

7x – y- 42 = 0

We get,

Y = 7x â€“ 42

Now, substituting x = 0 in y = 7x â€”42,

we get y = -42

Substituting x = 6 in y = 7x – 42,

we get y = 0

X | 0 | 6 |

Y | -42 | 0 |

We are given,

3x â€“ y + 6 =0

We get,

Y = 3x + 6

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

We get y = 6

Substituting x= â€”2 in y = 3x + 6,

We get y = 0

X | 0 | -2 |

Y | 6 | 0 |

The red -line represents the equation 7xâ€”y-42 =0.

The blue-line represents the equation 3x â€”y +6 =0.

*Q21: Â 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 Â (i) \(2\frac{1}{2}\) Hours (ii) \(\frac{1}{2}\) Hour*

**A21:**

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

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