In each of the following cases,
Question:

In each of the following cases, find a and b.

(i) (2a + b, a – b) = (8, 3)

(ii) (a/4 , a – 2b) = (0, 6 + b)

Solution:

(i)

According to the question,

(2a + b, a – b) = (8, 3)

Given the ordered pairs are equal, so corresponding elements will be equal.

Hence,

2a + b = 8 and a–b = 3

Now a–b = 3

⇒a = 3 + b

Substituting the value of a in the equation 2a + b = 8,

We get,

2(3 + b) + b = 8

⇒ 6 + 2b + b = 8

⇒ 3b = 8–6 = 2

⇒ b = 2/3

Substituting the value of b in equation (a–b = 3),

We get,

⇒ a – 2/3 = 3

⇒ a = 3 + 2/3

⇒ a = (9 + 2)/3

⇒ a = 11/3

Hence the value of a = 11/3 and b = 2/3 respectively.

(ii)

According to the question,

$\left(\frac{a}{4}, a-2 b\right)=(0,6+b)$

Given the ordered pairs are equal, so corresponding elements will be equal.

a/4 = 0 and a – 2b = 6 + b

Now a/4 = 0

⇒a = 0

Substituting the value of a in the equation (a–2b = 6 + b),

We get,

0 – 2b = 6 + b

⇒ – 2b – b = 6

⇒ – 3b = 6

⇒ b = – 6/3

⇒ b = – 2

Hence, the value of a = 0 and b = – 2 respectively

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