# 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 MCQ

Correct option: (b)

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

Correct option: (a)

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

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.

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.

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.

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.

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.

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).

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).

Correct option: (c)

Since, the y coordinate is 3, the graph of the line y = 3 passes through the point (2, 3).

Correct option: (d)

Infinitely many linear equations can be satisfied by x = 2 and y = 3.

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

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.

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

## Chapter 4 - Linear Equations in Two Variables Ex. 4B

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.

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

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

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.

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

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

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

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

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.

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.

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.

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.

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.

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).

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).

**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).

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).

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

**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

**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

**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).

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.

## Chapter 4 - Linear Equations in Two Variables Ex. 4A

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_{ }

On comparing this equation with ax + by + c = 0, we obtain

a = 10, b = -1 and c = 30

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

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

⇒ 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

On comparing this equation with ax + by + c = 0, we obtain

a = , b = and c = -5

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

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

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

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

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

We have,

⇒ 3x - 8y - 1 = 0

On comparing this equation with ax + by + c = 0, we obtain

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

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.

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.

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.

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.

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.

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.

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.

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.

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

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

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

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