Question

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

• A. $t=1$
• B. $t=-2$
• C. $t=2$
• D. $t=0$

Answer 1

The correct option is D $t=0$

Given, $\frac{t+2}{3}+\frac{1}{t+1}=\frac{t+3}{2}-\frac{t-1}{6}$
$\Rightarrow \frac{(t+1)(t+2)+3}{3(t+1)}=\frac{6(t+3)-2(t-1)}{12}$
$\Rightarrow \frac{t^2+2t+t+2+3}{3(t+1)}=\frac{6t+18-2t+2}{12}$
$\Rightarrow 4(t^2+3t+5)=(4t+20)(t+1)$
$\Rightarrow 4(t^2+3t+5)=4(t+5)(t+1)$
$\Rightarrow t^2+3t+5=t^2+5t+t+5$
$\Rightarrow 3t=0$
$\Rightarrow t=0$