Question

Solve: $\frac{t+2}{3}+\frac{1}{t+1}=\frac{t+3}{2}-\frac{t-1}{6}$.

• A. $t=\frac{1}{2}$
• B. $t=-2$
• C. $t=2$
• D. $t=0$

Answer 1

The correct option is A $t=\frac{1}{2}$

Given, $\frac{t+2}{3}+\frac{1}{t+1}=\frac{t+3}{2}-\frac{t-1}{6}$.

Transposing $\frac{1}{t+1}$ to the RHS, we get,

$\Rightarrow \frac{1}{t+1}=\frac{t+3}{2}-\frac{t-1}{6}-\frac{t+2}{3}$

$\Rightarrow \frac{1}{t+1}=\frac{3(t+3)-(t-1)-2(t+3)}{6}$...[Taking LCM of the denominators on the RHS]

$\Rightarrow \frac{1}{t+1}=\frac{3t+9-t+1-2t-6}{6}$ ...[By Distribution Law]

$\Rightarrow \frac{1}{t+1}=\frac{3t-3t+10-6}{6}$

$\Rightarrow \frac{1}{t+1}=\frac{4}{6}$

$\Rightarrow \frac{1}{t+1}=\frac{2}{3}$ ...[Cross-multiplying the denominators]

$\Rightarrow 3=2(t+1)$

$\Rightarrow 3=2t+2$...[By Distribution Law].

Transposing $x$ terms to one side, we get,

$\Rightarrow 2t=3-2=1$

$\therefore t=\frac{1}{2}$.

Hence, option $A$ is correct.