Question

Solve the given pair of equations by substitution/cross multiplication method

(i) 11+15+23=0;7−2−20=0.

(ii) +=5;3+2=13;≠0,≠0.​

Answer 1

Answer:

1) The given system of equations is:

kx + 2y – 5 = 0

3x + y – 1 = 0

The above equations are of the form

a1x + b1y − c1 = 0

a2x + b2y − c2 = 0

Here, a1 = k, b1 = 2, c1 = −5

a2 = 3, b2 = 1, c2 = -1

So according to the question,

For unique solution, the condition is

a1 / a2 ≠ b1 / b2

k/3 ≠ 2/1

⇒ k ≠ 6

2) Let's assume the cost of a pen and pencil be ₹ x and ₹ y respectively.

Then, forming equations according to the question

5x + 6y = 9 … (i)

3x + 2y = 5 … (ii)

On multiplying equation (i) by 2 and equation (ii) by 6, we get

10x + 12y = 18 … (iii)

18x + 12y = 30 … (iv)

Now on subtracting equation (iii) from equation (iv), we get

18x – 10x + 12y – 12y = 30 – 18

8x = 12

x = 3/2 = 1.5

Putting x = 1.5 in equation (i), we find y

5(1.5) + 6y = 9

6y = 9 – 7.5

y = (1.5)/ 6 = 0.25

Therefore, the cost of one pen = ₹ 1.50 and so the cost of one pencil = ₹ 0.25

Step-by-step explanation:

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