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Given, (23x + 1 × 42y + 1 = 2z – x). If (x + y) = 1, find the value of z.1). 32). 53). 74). 9 |
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Answer» ⇒ 23x + 1 × 22(2y + 1) = 2z – x ⇒ 2(3x + 1 + 4y + 2) = 2z – x Equating the powers of 2, ⇒ 3x + 4y + 3 = z – x ⇒ z = 4x + 4y + 3 ⇒ z = 4(x + y) + 3 Putting (x + y) = 1, ∴ z = 4 + 3 = 7 |
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