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

## Linear Equations in Two Variables Exercise Ex. 7.1

### Solution 1

### Solution 2

### Solution 3

## Linear Equations in Two Variables Exercise Ex. 7.2

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7(i)

### Solution 7(ii)

### Solution 7(iii)

## Linear Equations in Two Variables Exercise Ex. 7.3

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 1(v)

### Solution 1(vi)

### Solution 1(vii)

### Solution 1(viii)

### Solution 2

### Solution 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 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9(i)

### Solution 9(ii)

### Solution 9(iii)

### Solution 9(iv)

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### 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:**

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

## Linear Equations in Two Variables Exercise Ex. 7.4

### Solution 1(i)

### 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 = 3Point A represents 3 on number line. On Cartesian plane, equation represents all points on x axis for which y = 3**

### Solution 1(iv)

### Solution 1(v)

### Solution 2(i)

### Solution 2(ii)

### Solution 3(i)

### 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**

### Solution 5

## 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).

### Solution 10