#### 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

⇒ 5*x *= 9*x *+ 160

⇒5*x *– 9*x *= 160 ⇒ – 4*x
*= 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 equation**mx – 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