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Class Nine Maths Chapter 4 Additional Q-A NCERT

Chapter 4: Linear Equations in Two Variables Additional Questions- Answers

Short Answer Questions Type:         2 Marks

 1. The linear equation F =( 9/5) C + 32 is used to convert the temperature from Fahrenheit to Celsius or vice-versa. Find the temperature which is numerically same in both Fahrenheit and Celsius.

 Sol. Given that F =( 9/5) C + 32

 On putting F = C = x, we get x =( 9/5) x + 32

 ⇒ 5x = 9x + 160

⇒5x – 9x = 160 ⇒ – 4x = 160

x = – 40

2. At what point does the graph of the linear equation x + y = 5 meet a line which is parallel to the y-axis, at a distance of 2 units from the origin in the positive direction of x-axis ?

Sol. The coordinates of the points lying on the line parallel to the y-axis, at a distance 2 units from the origin and in the positive direction of the x-axis are of the form (2, a). Putting x = 2, y = a in the equation x + y = 5, we get a = 3. Hence, the required point is (2, 3).

3. Draw the graph of the equation represented by a straight line which is parallel to the x-axis and at a distance of 3 units below it.

Sol. We know that any straight line parallel to the x-axis is given by y = k, where k is the distance of the line from the x-axis. Here, k = – 3. Therefore, the equation of the line is y = – 3. To draw the graph of this equation, plot the points (1, – 3) and (2, – 3) and join them. This is the required graph.

4. If x = –1, y = 4 is a solution of the equationmx – y = – 6, then find the value of m.

Sol. x = – 1 and y = 4 is a solution of the equation mx – y = – 6

So, m (– 1) – 4 = – 6

⇒ – m = – 6 + 4 = – 2 ⇒ m = 2

5. Find the solution of 4x – y = 5.

Sol. For, x = 2, y = 3,

LHS = 4 × 2 – 3 = 5 = RHS

For, x = 3, y = 7, LHS = 4 × 3 – 7 = 5 = RHS

For, x = 4, y = 11, LHS = 4 × 4 – 11 = 5 = RHS.

6. Find the solution of x + y – 7x

Sol. Substituting x = 5 and y = 2 in LHS of x + y – 7 = 0, we get

5 + 2 – 7 = 0 ⇒ 7 – 7 = 0 = R HS

Hence, x = 5 and y = 2 is the solution of x + y = 7 .

7. Find the value of p from the equation 3x + 4y = p, if its one solution is x = 2, y = 1.

Sol. Substituting x = 2 and y = 1 in equation 3x + 4y = p, we get

3 × 2 + 4 × 1 = p ⇒ 6 + 4 = p ⇒ p = 10.

8. Find whether the given ordered pair is a solution of the given linear equation :

(i) 2x – 4y = 32; (8, – 4) (ii) 4x – 2y = 10; (3, – 1)

Sol. (i) The given equation is 2x – 4y – 32 = 0

Putting x = 8 and y = – 4 , we get

LHS = 2 × 8 – 4 × (– 4) – 32

= 16 + 16 – 32 = 0 = RHS

Since, LHS = RHS, therefore, (8, – 4) is the solution of the given equation.

(ii) The given equation is 4x – 2y – 10 = 0

Putting x = 3 and y = – 1, we get

LHS = 4 × 3 – 2 × (– 1) – 10 = 12 + 2 – 10 _ 0

Since, LHS _ RHS, therefore, (3, – 1) is not a solution of the given equation.

9. For what value of p, the linear equation 2x + py = 8 has equal values of x and y for its solution?

Sol. We have, 2x + py = 8 …. (i)

For equal values of x and y put x = y in equation (i)

2x + px = 8 ⇒ px = 8 – 2x


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

(i) 5x = – y (ii) y = – 4x

Sol. (i) 5x = – y can be written as

5x + y + 0 = 0

∴ a = 5, b = 1 and c = 0

(ii) y = – 4x can be written as

y + 4x + 0 = 0

⇒ 4x + y + 0 = 0

∴ a = 4, b = 1 and c = 0

11. Frame a linear equation in the form ax + by + c = 0 by using the given values of a, b and c.

(i) a = – 2, b = 3, c = 4

(ii) a = 5, b = 0, c = 7

Sol. (i) The given equation is ax + by + c = 0

Putting a = – 2, b = 3 and c = 4, we get, – 2x + 3y + 4 = 0, which is the required linear equation.

(ii) The given equation is ax + by + c = 0

Putting a = 5, b = 0 and c = 7, we get 5x + 0y + 7 = 0, which is the required linear equation.

12. Write each of the following equations in the form of ax + by + c = 0. Also, write the values of a, b and c.

(i) 4 = 5x – 3y (ii) x – 4 = 3 y

Sol. (i) The given equation is

4 = 5x – 3y

⇒ 5x – 3y = 4

⇒ 5x – 3y – 4 = 0

a = 5, b = – 3, c = –4.

(ii) The given equation is

x – 4 = 3 y

⇒ x – 3 y – 4 = 0

a = 1, b = – 3 , c = –4.

13. The cost of a book is four times the cost of a notebook. Express this statement as linear equation in two variables.

Sol. Let the cost of a book be Rs x.

and the cost of a notebook Rs y.

According to condition,

The cost of a book = 4 × The cost of a notebook

⇒ x = 4 × y

⇒ x = 4y

⇒ x – 4y = 0

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