# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables

## Chapter 4 – Linear Equations in Two Variables Exercise Ex. 4A

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in

each case

3x + 5y = 7.5

We

have,

3x + 5y = 7.5

⇒ 3x + 5y –

7.5 = 0

⇒ 6x + 10y

– 15 = 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 6, b = 10 and c = -15_{ }

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in

each case

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 10, b = -1 and c = 30

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in

each case

3y – 2x = 6

We

have,

3y – 2x = 6

⇒ -2x + 3y – 6

= 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= -2, b = 3 and c =

-6

each case

4x = 5y

We

have,

4x = 5y

⇒

4x – 5y = 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 4, b = -5 and c = 0

each case

⇒ 6x – 5y = 30

⇒ 6x – 5y – 30 = 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 6, b = -5 and c = –30

each case

On comparing this equation with ax

+ by + c = 0, we obtain

a

= , b = and c = -5

each case

x = 6

We

have,

x

= 6

⇒ x – 6 = 0

⇒ 1x + 0y – 6 =

0

⇒ x + 0y – 6 = 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 1, b = 0 and c = -6

each case

3x – y = x – 1

We

have,

3x – y = x – 1

⇒ 3x – x – y + 1

= 0

⇒ 2x – y + 1 = 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 2, b = -1 and c = 1

each case

2x + 9 = 0

We

have,

2x + 9 = 0

⇒ 2x + 0y + 9 =

0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 2, b = 0 and c = 9

each case

4y = 7

We

have,

4y = 7

⇒ 0x + 4y – 7 = 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 0, b = 4 and c = -7

each case

x + y = 4

We

have,

x + y = 4

⇒ x + y – 4 = 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 1, b = 1 and c = -4

each case

We

have,

⇒ 3x – 8y – 1 = 0

On comparing this equation with ax

+ by + c = 0, we obtain

a

= 3, b = -8 and c = -1

Check which of the following are the solutions of the

equation 5x – 4y = 20.

(4, 0)

Given

equation is 5x – 4y = 20

Substituting

x = 4 and y = 0 in L.H.S. of given equation, we get

L.H.S.

= 5x – 4y

= 5(4) – 4(0)

= 20 – 0

= 20

= R.H.S.

Hence,

(4, 0) is the solution of the given equation.

Check which of the following are the solutions of the

equation 5x – 4y = 20.

(0, 5)

Given

equation is 5x – 4y = 20

Substituting

x = 0 and y = 5 in L.H.S. of given equation, we get

L.H.S.

= 5x – 4y

= 5(0) – 4(5)

= 0 – 20

= -20

≠ R.H.S.

Hence,

(0, 5) is not the solution of the given equation.

Check which of the following are the solutions of the

equation 5x – 4y = 20.

Given

equation is 5x – 4y = 20

Substituting

x = -2 and y = in L.H.S. of given

equation, we get

L.H.S.

= 5x – 4y

= 5(-2) – 4

= -10 – 10

= -20

≠ R.H.S.

Hence,

is not the

solution of the given equation.

Check which of the following are the solutions of the

equation 5x – 4y = 20.

(0, -5)

Given

equation is 5x – 4y = 20

Substituting

x = 0 and y = -5 in L.H.S. of given equation, we

get

L.H.S.

= 5x – 4y

= 5(0) – 4(-5)

= 0 + 20

= 20

= R.H.S.

Hence,

(0, -5) is the solution of the given

equation.

Check which of the following are the solutions of the

equation 5x – 4y = 20.

Given

equation is 5x – 4y = 20

Substituting

x = 2 and y = in L.H.S. of given

equation, we get

L.H.S.

= 5x – 4y

= 5(2) – 4

= 10 + 10

= 20

= R.H.S.

Hence,

is the

solution of the given equation.

Find five different solutions of each of the following

equations:

2x – 3y = 6

Given

equation is 2x – 3y = 6

Substituting

x = 0 in the given equation, we get

2(0)

– 3y = 6

⇒ 0 – 3y = 6

⇒ 3y = -6

⇒ y = -2

So,

(0, -2) is the solution of the given equation.

Substituting

y = 0 in the given equation, we get

2x

– 3(0) = 6

⇒ 2x – 0 = 6

⇒ 2x = 6

⇒ x = 3

So,

(3, 0) is the solution of the given equation.

Substituting

x = 6 in the given equation, we get

2(6)

– 3y = 6

⇒ 12 – 3y = 6

⇒ 3y = 6

⇒ y = 2

So,

(6, 2) is the solution of the given equation.

Substituting

y = 4 in the given equation, we get

2x

– 3(4) = 6

⇒ 2x – 12 = 6

⇒ 2x = 18

⇒ x = 9

So,

(9, 4) is the solution of the given equation.

Substituting

x = -3 in the given equation, we get

2(-3)

– 3y = 6

⇒ -6 – 3y = 6

⇒ 3y = -12

⇒ y = -4

So,

(-3, -4) is the solution of the given equation.

Find five different solutions of each of the following

equations:

Given

equation is

Substituting

x = 0 in (i), we get

4(0)

+ 3y = 30

⇒ 3y = 30

⇒ y = 10

So,

(0, 10) is the solution of the given equation.

Substituting

x = 3 in (i), we get

4(3)

+ 3y = 30

⇒ 12 + 3y = 30

⇒ 3y = 18

⇒ y = 6

So,

(3, 6) is the solution of the given equation.

Substituting

x = -3 in (i), we get

4(-3)

+ 3y = 30

⇒ -12 + 3y = 30

⇒ 3y = 42

⇒ y = 14

So,

(-3, 14) is the solution of the given equation.

Substituting

y = 2 in (i), we get

4x

+ 3(2) = 30

⇒ 4x + 6 = 30

⇒ 4x = 24

⇒ x = 6

So,

(6, 2) is the solution of the given equation.

Substituting

y = -2 in (i), we get

4x

+ 3(-2) = 30

⇒ 4x – 6 = 30

⇒ 4x = 36

⇒ x = 9

So,

(9, -2) is the solution of the given equation.

Find five different solutions of each of the following

equations:

3y = 4x

Given

equation is 3y = 4x

Substituting

x = 3 in the given equation, we get

3y

= 4(3)

⇒ 3y = 12

⇒ y = 4

So,

(3, 4) is the solution of the given equation.

Substituting

x = -3 in the given equation, we get

3y

= 4(-3)

⇒ 3y = -12

⇒ y = -4

So,

(-3, -4) is the solution of the given equation.

Substituting

x = 9 in the given equation, we get

3y

= 4(9)

⇒ 3y = 36

⇒ y = 12

So,

(9, 12) is the solution of the given equation.

Substituting

y = 8 in the given equation, we get

3(8)

= 4x

⇒ 4x = 24

⇒ x = 6

So,

(6, 8) is the solution of the given equation.

Substituting

y = -8 in the given equation, we get

3(-8)

= 4x

⇒ 4x = -24

⇒ x = -6

So,

(-6, -8) is the solution of the given equation.

If x = 3 and y = 4 is a solution of the equation 5x – 3y =

k, find the value of k.

Since x = 3 and y = 4 is a

solution of the equation 5x – 3y = k, substituting x = 3 and y = 4 in

equation 5x – 3y = k, we get

5(3)

– 3(4) = k

⇒ 15 – 12 = k

⇒ k = 3

If x = 3k + 2 and y = 2k – 1 is a solution of the equation

4x – 3y + 1 = 0, find the value of k.

Since x = 3k + 2 and y = 2k – 1 is

a solution of the equation 4x – 3y + 1 = 0, substituting these values in

equation, we get

4(3k + 2) – 3(2k – 1) + 1 = 0

⇒ 12k + 8 – 6k + 3 + 1 = 0

⇒ 6k + 12 = 0

⇒ 6k = -12

⇒ k = -2

The cost of 5 pencils is equal to the cost of 2 ballpoints.

Write a linear equation in two variables to represent this statement. (Take

the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y).

Let

the cost of one pencil be Rs. x and that of one

ballpoint be Rs. y.

Then,

Cost

of 5 pencils = Rs. 5x

Cost

of 2 ballpoints = Rs. 2y

According

to given statement, we have

5x

= 2y

⇒ 5x – 2y = 0

## Chapter 4 – Linear Equations in Two Variables Exercise Ex. 4B

Draw the graph of each of the following equation.

x = 4

x

= 4 is a line parallel to the Y-axis, at a distance of 4 units from it, to

its right.

Draw the graph of each of the following equation.

x + 4 = 0

x

+ 4 = 0

⇒ x = -4, which

is a line parallel to the Y-axis, at a distance of 4 units from it, to its

left.

Draw the graph of each of the following equation.

y = 3

y

= 3 is a line parallel to the X-axis, at a distance of 3 units from it, above

the X-axis.

Draw the graph of each of the following equation.

y = -3

y

= -3 is a line parallel to the X-axis, at a distance of 3 units from it,

below the X-axis.

Draw the graph of each of the following equation.

x = -2

x

= -2 is a line parallel to the Y-axis, at a distance of 2 units from it, to

its left.

Draw the graph of each of the following equation.

x = 5

x

= 5 is a line parallel to the Y-axis, at a distance of 5 units from it, to

its right.

Draw the graph of each of the following equation.

y + 5 = 0

y

+ 5 = 0

⇒ y = -5, which

is a line parallel to the X-axis, at a distance of 5 units from it, below the

X-axis.

Draw the graph of each of the following equation.

y = 4

y

= 4 is a line parallel to the X-axis, at a distance of 4 units from it, above

the X-axis.

Draw the graphs of the lines x – y = 1 and 2x + y = 8.

Shade the area formed by these two lines and the y-axis. Also, find this

area.

**Graph of the equation x – y = 1**

⇒ y = x – 1

When

x = 1, then y = 1 – 1 = 0

When

x = 2, then y = 2 – 1 = 1

Thus,

we have the following table:

x | 1 | 2 |

y | 0 | 1 |

Now,

plot the points A(1, 0) and B(2, 1) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of x – y = 1.

**Graph of the equation 2x + y = 8**

⇒ y = 8 – 2x

When

x = 2, then y = 8 – 2(2) = 8 – 4 = 4

When

x = 3, then y = 8 – 2(3) = 8 – 6 = 2

Thus,

we have the following table:

x | 2 | 3 |

y | 4 | 2 |

Now,

plot the points C(2, 4) and D(3, 2) on a graph

paper.

Join

CD and extend it in both the directions.

Then,

the line CD is the required graph of 2x + y = 8.

The

two graph lines intersect at point D(3, 2).

The

area enclosed by the lines and Y-axis is shown in the graph.

Draw

DM perpendicular from D on Y-axis.

DM

= x-coordinate of point D(3, 2) = 3

And,

EF = 9

Area

of shaded region = Area of ΔDEF

Draw the graph for each of the equations x + y = 6 and x –

y = 2 on the same graph paper and find the coordinates of the point where the

two straight lines intersect.

*Back answer incorrect.

**Graph of the equation x + y = 6**

⇒ y = 6 – x

When

x = 2, then y = 6 – 2 = 4

When

x = 3, then y = 6 – 3 = 3

Thus,

we have the following table:

x | 2 | 3 |

y | 4 | 3 |

Now,

plot the points A(2, 4) and B(3, 3) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of x + y = 6.

**Graph of the equation x – y = 2**

⇒ y = x – 2

When

x = 3, then y = 3 – 2 = 1

When

x = 4, then y = 4 – 2 = 2

Thus,

we have the following table:

x | 3 | 4 |

y | 1 | 2 |

Now,

plot the points C(3, 1) and D(4, 2) on a graph

paper.

Join

CD and extend it in both the directions.

Then,

the line CD is the required graph of x – y = 2.

The

two graph lines intersect at point D(4, 2).

Two students A and B contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the

earthquake victims. Write a linear equation to satisfy the above data and

draw its graph.

Let

the amount contributed by students A and B be Rs. x

and Rs. y respectively.

Total

contribution = 100

⇒** x + y = 100**

⇒ y = 100 – x

When

x = 25, then y = 100 – 25 = 75

When

x = 50, then y = 100 – 50 = 50

Thus,

we have the following table:

x | 25 | 50 |

y | 75 | 50 |

Now,

plot the points A(25, 75) and B(50, 50) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of x + y = 100.

Draw the graph of the equation y = 3x.

From your graph, find the value of y when x = 2.

y

= 3x

When

x = 1, then y = 3(1) = 3

When

x = -1, then y = 3(-1) = -3

Thus,

we have the following table:

x | 1 | -1 |

y | 3 | -3 |

Now,

plot the points A(1, 3) and B(-1, -3) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of y = 3x.

Reading

the graph

Given:

x = 2. Take a point M on the X-axis such that OM = 2.

Draw

MP parallel to the Y-axis, cutting the line AB at P.

Clearly,

PM = 6

Thus,

when x = 2, then y = 6.

Draw

the graph of the equation y = 3x. From your graph, find the value of y when x

= -2.

The

given equation is y = 3x.

Putting

x = 1, y = 3 1 = 3

Putting

x = 2, y = 3 2 = 6

Thus,

we have the following table:

x | 1 | 2 |

y | 3 | 6 |

Plot

points (1,3) and (2,6) on a graph paper and join

them to get the required graph.

Take

a point P on the left of y-axis such that the distance of point P from the

y-axis is 2 units.

Draw

PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to

x-axis meeting y-axis at N.

So,

y = ON = -6.

Draw the graph of the equation x + 2y – 3 = 0. From your graph, find the value of y when x = 5.

The given equation is,

x + 2y – 3 = 0

x = 3 – 2y

Putting y = 1,x = 3 – (2 1) = 1

Putting y = 0,x = 3 – (2 0) = 3

Thus, we have the following table:

x | 1 | 3 |

y | 1 | 0 |

Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.

Take a point Q on x-axis such that OQ = 5.

Draw QP parallel to y-axis meeting the line (x = 3 – 2y) at P.

Through P, draw PM parallel to x-axis cutting y-axis at M.

So, y = OM = -1.

Draw the graph of the equation x + 2y – 3 = 0.

From your graph, find the value of y when x = -5

x

+ 2y – 3 = 0

⇒ 2y = 3 – x

When

x = -1, then

When

x = 1, then

Thus,

we have the following table:

x | -1 | 1 |

y | 2 | 1 |

Now,

plot the points A(-1, 2) and B(1, 1) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of x + 2y – 3 = 0.

Reading

the graph

Given:

x = -5. Take a point M on the X-axis such that OM = -5.

Draw

MP parallel to the Y-axis, cutting the line AB at P.

Clearly,

PM = 4

Thus,

when x = -5, then y = 4.

Draw

the graph of the equation 2x – 3y = 5. From the graph, find (i) the value of y when x = 4, and (ii) the value of x

when y = 3.

The given equation is, 2x – 3y = 5

_{}

Now, if x = 4, then

_{}

And, if x = -2, then

_{}

Thus, we have the following table:

x | 4 | -2 |

y | 1 | -3 |

Plot points (4,1)

and (-2,-3) on a graph paper and join them to get the required graph.

(i) When

x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to

its right cutting the line at Q and through Q draw a line parallel to x-axis

cutting y-axis which is found to be at a distance of 1 units above x-axis.

Thus, y = 1 when x = 4.

(ii) When y = 3, draw a line parallel to

x-axis at a distance of 3 units from x-axis and above it, cutting the line at

point P. Through P, draw a line parallel to y-axis meeting x-axis at a point

which is found be 7 units to the right of y axis.

Thus, when y = 3, x = 7.

Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.

The given equation is 2x + y = 6

y = 6 – 2x

Now, if x = 1, then y = 6 – 2 1 = 4

And, if x = 2, then y = 6 – 2 2 = 2

Thus, we have the following table:

x | 1 | 2 |

y | 4 | 2 |

Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.

We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.

So, the co-ordinates of P are (3,0).

Draw

the graph of the equation 3x + 2y = 6. Find the coordinates of the point,

where the graph cuts the y-axis.

The given equation is 3x + 2y = 6

2y = 6 – 3x

Now, if x = 2, then

_{}

And, if x = 4, then

_{}

Thus, we have the following table:

x | 2 | 4 |

y | 0 | -3 |

Plot points (2, 0) and (4,-3) on a

graph paper and join them to get the required graph.

We find that the line 3x + 2y = 6

cuts the y-axis at a point P which is 3 units above the x-axis.

So, co-ordinates of P are (0,3).

Draw the graphs of the equations 3x – 2y = 4 and x + y – 3

= 0. On the same graph paper, find the coordinates of the point where the two

graph lines intersect.

**Graph of the equation 3x – 2y = 4**

⇒ 2y = 3x – 4

When

x = 2, then

When

x = -2, then

Thus,

we have the following table:

x | 2 | -2 |

y | 1 | -5 |

Now,

plot the points A(2, 1) and B(-2, -5) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of 3x – 2y = 4.

**Graph of the equation x + y – 3 = 0**

⇒ y = 3 – x

When

x = 1, then y = 3 – 1 = 2

When

x = -1, then y = 3 – (-1) = 4

Thus,

we have the following table:

x | 1 | -1 |

y | 2 | 4 |

Now,

plot the points C(1, 2) and D(-1, 4) on a graph

paper.

Join

CD and extend it in both the directions.

Then,

the line CD is the required graph of x + y – 3 = 0.

The

two graph lines intersect at point A(2, 1).

Draw the graph of the line 4x + 3y = 24.

Write the coordinates of the points where this line

intersects the x-axis and the y-axis.

4x

+ 3y = 24

⇒ 3y = 24 – 4x

When

x = 0, then

When

x = 3, then

Thus,

we have the following table:

x | 0 | 3 |

y | 8 | 4 |

Now,

plot the points A(0, 8) and B(3, 4) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of 4x + 3y = 24.

Reading

the graph

The

graph of line 4x + 3y = 24 intersects the X-axis at point C(6,

0) and the Y-axis at point A(0, 8).

Draw the graph of the line 4x + 3y = 24.

Use this graph to find the area of the triangle formed by

the graph line and the coordinate axes.

4x

+ 3y = 24

⇒ 3y = 24 – 4x

When

x = 0, then

When

x = 3, then

Thus,

we have the following table:

x | 0 | 3 |

y | 8 | 4 |

Now,

plot the points A(0, 8) and B(3, 4) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of 4x + 3y = 24.

Reading

the graph

Required

area = Area of ΔAOC

Draw the graphs of the lines 2x + y = 6 and 2x – y + 2 =

0. Shade the region bounded by these two lines and the x-axis. Find the area

of the shaded region.

**Graph of the equation 2x + y = 6**

⇒ y = 6 – 2x

When

x = 1, then y = 6 – 2(1) = 6 – 2 = 4

When

x = 2, then y = 6 – 2(2) = 6 – 4 = 2

Thus,

we have the following table:

x | 1 | 2 |

y | 4 | 2 |

Now,

plot the points A(1, 4) and B(2, 2) on a graph

paper.

Join

AB and extend it in both the directions.

Then,

the line AB is the required graph of 2x + y = 6.

**Graph of the equation 2x – y + 2 = 0**

⇒ y = 2x + 2

When

x = -1, then y = 2(-1) + 2 = -2 + 2 = 0

When

x = 2, then y = 2(2) + 2 = 4 + 2 = 6

Thus,

we have the following table:

x | -1 | 2 |

y | 0 | 6 |

Now,

plot the points C(-1, 0) and D(2, 6) on a graph

paper.

Join

CD and extend it in both the directions.

Then,

the line CD is the required graph of 2x – y + 2 = 0.

The

two graph lines intersect at point A(1, 4).

The

area enclosed by the lines and X-axis is shown in the graph.

Draw

AM perpendicular from A on X-axis.

PM

= y-coordinate of point A(1, 4) = 4

And,

CP = 4

Area

of shaded region = Area of ΔACP

## Chapter 4 – Linear Equations in Two Variables Exercise MCQ

The equation of the x-axis is

(a) x

= 0

(b) y

= 0

(c) x

= y

(d) x

+ y = 0

Correct

option: (b)

The

equation of the x-axis is y = 0.

The graph of y + 2 = 0 is a line

(a) making

an intercept -2 on the x-axis

(b) making

an intercept -2 on the y-axis

(c) parallel

to the x-axis at a distance of 2 units below the x-axis

(d) parallel

to the y-axis at a distance of 2 units to the left of y-axis

Correct option: (c)

The graph of y + 2 = 0 is a line

parallel to the x-axis at a distance of 2 units below the x-axis.

The graph of the linear equation 2x + 3y = 6 meets the

y-axis at the point

(a) (2,

0)

(b) (3,

0)

(c) (0,

2)

(d) (0, 3)

Correct

option: (c)

When

a graph meets the y-axis, the x coordinate is zero.

Thus,

substituting x = 0 in the given equation, we get

2(0)

+ 3y = 6

⇒ 3y = 6

⇒ y = 2

Hence,

the required point is (0, 2).

The graph of the linear equation 2x + 5y = 10 meets the

x-axis at the point

(a) (0,

2)

(b) (2,

0)

(c) (5,

0)

(d) (0, 5)

Correct

option: (c)

When

a graph meets the x-axis, the y coordinate is zero.

Thus,

substituting y = 0 in the given equation, we get

2x

+ 5(0) = 10

⇒ 2x = 10

⇒ x = 5

Hence,

the required point is (5, 0).

The graph of the line x

= 3 passes through the point

(a) (0,3)

(b) (2,3)

(c) (3,2)

(d) None of these

The graph of the line y = 3 passes though the point

(a) (3,

0)

(b) (3,

2)

(c) (2,

3)

(d) none

of these

Correct

option: (c)

Since,

the y coordinate is 3, the graph of the line y = 3 passes through the point

(2, 3).

The graph of the line y

= -3 does not pass through the point

(a) (2,-3)

(b) (3,-3)

(c) (0,-3)

(d) (-3,2)

The graph of the linear

equation x-y=0 passes through the point

If each of (-2,2), (0,0) and (2,-2) is a solution of a linear equation in x and y, then the equation is

(a) x-y=0

(b) x+y=0

(c) -x+2y=0

(d) x – 2y=0

How many linear equations can be satisfied by x = 2 and y

= 3?

(a) only

one

(b) only

two

(c) only

three

(d) Infinitely

many

Correct option: (d)

Infinitely many linear equations can be satisfied by x = 2

and y = 3.

A linear equation in two

variable x and y is of the form ax+by+c=0, where

(a) a≠0, b≠0

(b) a≠0, b=0

(c) a=0, b≠0

(d) a= 0, c=0

The equation of the y-axis is

(a) x

= 0

(b) y

= 0

(c) x

= y

(d) x

+ y = 0

Correct

option: (a)

The

equation of the y-axis is x = 0.

If (2, 0) is a solution of the linear equation 2x + 3y = k

then the value of k is

(a) 6

(b) 5

(c) 2

(d) 4

Correct option: (d)

Since, (2, 0) is a solution of the

linear equation 2x + 3y = k, substituting x = 2 and y = 0 in the given

equation, we have

2(2) + 3(0) = k

⇒

4 + 0 = k

⇒

k = 4

Any point on x-axis is

of the form:

(a) (x,y), where x ≠0 and y ≠0

(b) (0,y), where y ≠0

(c) (x,0), where x ≠0

(d) (y,y), where y ≠0

Any point on y-axis is

of the form

(a) (x,0), where x ≠ 0

(b) (0,y), where y ≠ 0

(c) (x,x), where x ≠ 0

(d) None of these

x = 5, y = 2 is a solution of the linear equation

(a) x

+ 2y = 7

(b) 5x

+ 2y = 7

(c) x

+ y = 7

(d) 5x + y = 7

Correct option: (c)

Substituting x = 5 and y = 2 in

L.H.S. of equation x + y = 7, we get

L.H.S. = 5 + 2 = 7 = R.H.S.

Hence, x = 5 and y = 2 is a

solution of the linear equation x + y = 7.

If the point (3, 4) lies on the graph of 3y = ax + 7 then

the value of a is

(a)

(b)

(c)

(d)

Correct option: (b)

Since the point (3, 4) lies on the

graph of 3y = ax + 7, substituting x = 3 and y = 4 in the given equation, we

get

3(4) = a(3) + 7

⇒

12 = 3a + 7

⇒

3a = 5

The point of the form (a,a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y = x

(d) the line x + y

= 0

The point of the form

(a,-a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y-x=0

(d) the line x + y

= 0

The linear equation 3x –

5y = 15 has

(a) a unique

solution

(b) two solutions

(c) infinitely many

solutions

(d) no solution

The equation 2x + 5y = 7 has a unique solution, if x and y

are

(a) natural

numbers

(b) rational

numbers

(c) positive

real numbers

(d) real

numbers

Correct

option: (a)

The equation 2x + 5y = 7 has a unique solution, if x and y

are natural numbers.

If we take x = 1 and y = 1, the given equation is

satisfied.

The graph of y = 5 is a line

(a) making

an intercept 5 on the x-axis

(b) making

an intercept 5 on the y-axis

(c) parallel

to the x-axis at a distance of 5 units from the origin

(d) parallel

to the y-axis at a distance of 5 units from the origin

Correct

option: (c)

The graph of y = 5 is a line

parallel to the x-axis at a distance of 5 units from the origin.

The graph of x = 4 is a line

(a) making

an intercept 4 on the x-axis

(b) making

an intercept 4 on the y-axis

(c) parallel

to the x-axis at a distance of 4 units from the origin

(d) parallel

to the y-axis at a distance of 4 units from the origin

Correct

option: (d)

The graph of x = 4 is a line

parallel to the y-axis at a distance of 4 units from the origin.

The graph of x + 3 = 0 is a line

(a) making

an intercept -3 on the x-axis

(b) making

an intercept -3 on the y-axis

(c) parallel

to the y-axis at a distance of 3 units to the left of y-axis

(d) parallel

to the x-axis at a distance of 3 units below the x-axis

Correct option: (c)

The graph of x + 3 = 0 is a line parallel to the y-axis at

a distance of 3 units to the left of y-axis.