# Chapter 7 : Linear Equations in Two Variables - Rd Sharma Solutions for Class 9 Maths CBSE

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## Chapter 7 - Linear Equations in Two Variables Excercise Ex. 7.1

## Chapter 7 - Linear Equations in Two Variables Excercise Ex. 7.2

If x = 1 and y = 6 is solution of the equation 8x - ay + a^{2}= 0, find the value of a.

**Write two solutions of the form x = 0, y = a and x = b, y = 0 for the follwoing equation: 5x - 2y = 10**

**Write two solutions of the form x = 0, y = a and x = b, y = 0 for the following equation: -4x + 3y = 12**

## Chapter 7 - Linear Equations in Two Variables Excercise Ex. 7.3

**Plot the points (3,5) and (-1,3) on a graph paper and verify that the straight line passing through these points also passes through the point (1,4).**

**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).**

**From the choices given below, choose the equation whose graph is given in fig., ****(i) y = x****(ii) x + y = 0****(iii) y = 2x****(iv) 2 + 3y = 7x**** **

**From the choices given below, choose the equation whose graph is given in fig., ****(i) y = x + 2 ****(ii) y = x - 2****(iii) y = -x + 2****(iv) x + 2y = 6**** **

**Draw the graph of the equation 2x + 3y = 12. Find the graph, find the coordinates of the point. **** (i) whose y-coordinate is 3.****(ii) whose x-coordinate is -3**

**The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and ten's digit of the number are x and y respectively, then write the linear equation representing the above statement.**

**Draw the graph of y = |x|.**

**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|:**

**Draw the graph of y = |x| + 2.**

**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:**

**Ravish tells his daughter Aarushi, "Seven years ago, I was seven times as old as you were then. Also, three years form 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.**

## Chapter 7 - Linear Equations in Two Variables Excercise Ex. 7.4

**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**

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

**Give the geometrical representation of 2x + 13 = 0 as an equation in ****One variable**

**Give the geometrical representation of 2x + 13 = 0 as an equation in ****Two variables**

**Solve the equation 3x + 2 = x - 8, and represent the solution on (i) the number line.**

**Solve the equation 3x + 2 = x - 8, and represent the solution on (ii) the Cartesian plane.**

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

**Write the equation of the line that is parallel to x-axis and passing through the point ****(i) (0,3)****(ii) (0,-4)****(iii) (2,-5)****(iv) (3,4)**

**(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**

## Chapter 7 - Linear Equations in Two Variables Excercise 7.33

If (4, 19) is a solution of the equation y = ax + 3, then a =

(a) 3

(b) 4

(c) 5

(d) 6

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).

If (a, 4) lies on the graph of 3x + y = 10, then the value of a is

(a) 3

(b) 1

(c) 2

(d) 4

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).

The graph of the linear equation 2x - y = 4 cuts x-axis at

(a) (2, 0)

(b) (-2, 0)

(c) (0, -4)

(d) (0, 4)

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).

How many linear equations are satisfied by x = 2 and y = -3?

(a) Only one

(b) Two

(c) Three

(d) Infinitely many

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).

The equation x - 2 = 0 on number line is represented by

(a) a line

(b) a point

(c) infinitely many lines

(d) two lines

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).

x = 2, y = -1 is a solution of the linear equation

(a) x + 2y = 0

(b) x + 2y = 4

(c) 2x + y = 0

(d) 2x + y = 5

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).

If (2k - 1, k) is a solution of the equation 10x - 9y = 12, then k =

(a) 1

(b) 2

(c) 3

(d) 4

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).

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

(a) 1

(b) 2

(c) 3

(d) 5

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

= 2 - (-3)

= 2 + 3

= 5

Hence, correct option is (d).

The distance between the graphs of the equations y = -1 and y = 3 is

(a) 2

(b) 4

(c) 3

(d) 1

The distance between given two graphs

= 3 - (-1)

= 3 + 1

= 4

Hence, correct option is (b).

If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length

(a) 4 units

(b) 3 units

(c) 5 units

(d) none of these

## CBSE Class 9 Maths Homework Help

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