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# Class 9 RD SHARMA Solutions Maths Chapter 7 - Linear Equations in Two Variables

## Linear Equations in Two Variables Exercise Ex. 7.3

### Solution 4

The given points on the graph:

It is dear from the graph, the straight line passing through these points also passes through the point (1,4).

### Solution 15

We have,

y = |X|                               ...(i)

Putting x = 0, we get y = 0

Putting x = 2, we get y = 2

Putting x = -2, we get y = 2

Thus, we have the following table for the points on graph of |x|.

 x 0 2 -2 y 0 2 2

The graph of the equation y = |x|:

### Solution 16

We have,

y = |x| + 2                                                  ...(i)

Putting x = 0, we get y = 2

Putting x = 1, we get y = 3

Putting x = -1, we get y = 3

Thus, we have the following table for the points on graph of |x| + 2:

 x 0 1 -1 y 2 3 3

The graph of the equation y = |x| + 2:

## Linear Equations in Two Variables Exercise Ex. 7.4

### Solution 1(ii)

y + 3 = 0

y = -3

Point A represents -3 on number line.

On Cartesian plane, equation represents all points on x axis for which y = -3

### Solution 1(iii)

y = 3

Point A represents 3 on number line.

On Cartesian plane, equation represents all points on x axis for which y = 3

### Solution 3(ii)

On Cartesian plane, equation represents all points on y axis for which x = -5

### Solution 4

(i) The equation of the line that is parallel to x-axis and passing through the point (0,3) is y = 3

(ii) The equation of the line that is parallel to x-axis and passing through the point (0,-4) is y = -4

(iii) The equation of the line that is parallel to x-axis and passing through the point (2,-5) is y = -5

(iv) The equation of the line that is parallel to x-axis and passing through the point (3, 4) is y = 4

## Linear Equations in Two Variables Exercise 7.33

### Solution 1

y = ax + 3

If (4, 19) is its solution, then it must satisfy the equation.

Thus, we have

19 = a × 4 + 3

i.e. 4a = 16

i.e. a = 4

Hence, correct option is (b).

### Solution 2

3x + y = 10

If (a, 4) lies on its graph, then it must satisfy the equation.

Thus, we have

3(a) + 4 = 10

i.e. 3a = 6

i.e. a = 2

Hence, correct option is (c).

### Solution 3

On x-axis, the y-co-ordinate is always 0.

So, 2x - y = 4 will cut the x-axis where y = 0

i.e. 2x = 4

i.e. x = 2

Thus, 2x - y = 4 will cut the x-axis at (2, 0).

Hence, correct option is (a).

### Solution 4

From Point (2, -3) there are infinitely many lines passing in every-direction.

So (2, -3) is satisfied with infinite linear equations.

Hence, correct option is (d).

### Solution 5

Given equation is x – 2 = 0.

If this is treated as an equation in one variable x only, then it has the unique solution x = 2, which is a point on the number line.

However, when treated as an equation in two variables, it can be expressed as x - 2 = 0.

So as, an equation in two variables, x – 2 = 0 is represented by a single line parallel to y-axis at a distance of 2 units.

Hence, correct option is (a).

### Solution 6

Substituting x = 2 and y = -1 in the following equations:

L.H.S. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 = R.H.S.

L.H.S. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 ≠ 4 ≠ R.H.S.

L.H.S. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 ≠ 0 ≠ R.H.S.

L.H.S. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 ≠ 5 ≠ R.H.S.

Hence, correct option is (a).

### Solution 7

If (2k - 1, k) is solution of equation 10x - 9y = 12, then it must satisfy this equation.

Thus, we have

10(2k - 1) - 9k = 12

20k - 10 - 9k = 12

11k = 22

k = 2

Hence, correct option is (b).

### Solution 8

The distance between the graph of the equations x = -3 and x = 2

= 2 - (-3)

= 2 + 3

= 5

Hence, correct option is (d).

### Solution 9

The distance between given two graphs

= 3 - (-1)

= 3 + 1

= 4

Hence, correct option is (b).