t − (2t + 5) − 5(1 − 2t) = 2(3 + 4t) −3(t − 4)

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Solution

$\begin{array}{l} We have: \\ t - (2 t + 5) - 5 (1 - 2 t) = 2 (3 + 4 t) - 3 (t - 4) \\ \Rightarrow t - 2 t - 5 - 5 + 10 t =6+8t - 3 t + 12 \\ \Rightarrow t - 2 t + 10 t - 8 t + 3 t = 6 + 12 + 5 + 5 (By transposition) \\ \Rightarrow 14 t - 10 t =28 \\ \Rightarrow 4 t = 28 \\ \Rightarrow x = \frac{2 8^{7}}{4^{1}} \\ \Rightarrow x = 7 \\ CHECK: Substituting x =7 in the given equation, we get: \\ LHS= t - (2 t + 5) - 5 (1 - 2 t) \\ = t - 2 t - 5 - 5 + 10 t \\ =11 t - 2 t - 10 \\ =9 t - 10 \\ =9 \times 7 - 10 \\ =63 - 10 \\ =53 \\ RHS= 2 (3 + 4 t) - 3 (t - 4) \\ =6+8 t - 3 t + 12 \\ =5 t + 18 \\ =5 \times 7 + 18 \\ =35 + 18 \\ = 53 \\ ∴ LHS=RHS \\ Hence, x =7 is a solution of the given equation . \end{array}$

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t - (2 t + 5) - 5 (1 - 2 t) = 2 (3 + 4 t) - 3 (t - 4)

Subtract

$3 t^{4} - 4 t^{3} + 2 t^{2} - 6 t + 6 f r o m - 4 t^{4} + 8 t^{3} - 4 t^{2} - 2 t + 11$

Evaluate: $\frac{3 t - 2}{4} - \frac{2 t + 3}{3} = \frac{2}{3} - t$

Question 82 (iv)

Subtract

$3 t^{4} - 4 t^{3} + 2 t^{2} - 6 t + 6 f r o m - 4 t^{4} + 8 t^{3} - 4 t^{2} - 2 t + 11$

Solve:
$\frac{3 t - 2}{4} - \frac{2 t + 3}{3} = \frac{2}{3} - t$

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