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)
(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