Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables

Clear all your Class 9 doubts with EduRev Join for Free

Join for Free

Q1. Solve:

Hint:   ...(1)
    ...(2)

Sol. 

Subtracting (1) from (2),  = an – bm or
Thus,
Ans.

Q2. Solve : 141x + 103y = 217;   103x + 141y = 27

Sol. 141x + 103y = 217 .......... (i)

103x + 141y = 27......... (ii)

(i) × 141 - (ii) × 103

(141)²x + (103)(141)y = 217  ×  141

(130)²x + (103)(141)y = 27  ×  103

 Subtracting both equations, we get 

x= (217 x 141)-(27 x 103)/ 141²- 103²

^x=3

^{From equation (i)}

^{141 x 3+ 103 x y=217}

^{y= -2}

Q3. Cost of a pen is two and half times the cost of a pencil. Express this situation as a linear equation in two variables.

Sol. Let cost of a pen be ₹ x and cost of a pencil be ₹ y. According to statement of the question, we have x = 5/2 y ⇒ 2x = 5y or 2x – 5y = 0

Q4. Show that the points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7.

Sol.

We have the equation,

y = 9x – 7

For A (1, 2),

Substituting (x,y) = (1, 2),

We get,

2 = 9(1) – 7

2 = 9 – 7

2 = 2

For B (–1, –16),

Substituting (x,y) = (–1, –16),

We get,

–16 = 9(–1) – 7

-16 = – 9 – 7

-16 = – 16

For C (0, –7),

Substituting (x,y) = (0, –7),

We get,

– 7 = 9(0) – 7

-7 = 0 – 7

-7 = – 7

Hence, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.

Thus, A (1, 2), B (–1, –16) and C (0, –7) are solutions of the linear equation y = 9x – 7

Therefore, the points A (1, 2), B (–1, –16), C (0, –7) lie on the graph of linear equation y = 9x – 7.

Q5. Solve : x + y = 18; y + z = 12;    z + x = 16

Sol. 

x+y = 18 ; y+z = 12 ;z+x = 16.

x = 18-y ∴ z+18-y = 16 ⇒ y-z = z & y + z = 12

adding -

2y=14⇒y = 7 & z = 5

∴ x = 18-7 = 11

∴ x = 11, y = 7 & z = 5

Q6. Write four solutions for the following equation:

πx + y = 9

Sol. To find the four solutions of πx + y = 9 we substitute different values for x and y

Let x = 0

Then,

πx + y = 9

(π × 0)+y = 9

y = 9

(0,9)

Let x = 1

Then,

πx + y = 9

(π×1)+y = 9

π+y = 9

y = 9-π

(1,9-π)

Let y = 0

Then,

πx + y = 9

πx +0 = 9

πx = 9

x =9/π

(9/π,0)

Let x = -1

Then, πx + y = 9

(π(-1))+y = 9

-π + y = 9

y = 9+π

(-1,9+π)

The solutions are (0,9), (1,9-π),(9/π,0),(-1,9+π)