Question

Question 5
$\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$

Answer 1

$\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\Rightarrow \frac{3t-2}{4}-\frac{2t+3}{3}+t=\frac{2}{3}$
$\Rightarrow \frac{3(3t-2)-4(2t+3)+12t}{12}=\frac{2}{3}\Rightarrow \frac{9t-6-8t-12+12t}{12}=\frac{2}{3}$
$\Rightarrow \frac{13t-18}{12}=\frac{2}{3}\Rightarrow 3\times (13t-18)=2\times 12$
$\Rightarrow 39t-54=24\Rightarrow 39t=24+54$
$\Rightarrow 39t=78\Rightarrow t=\frac{78}{39}=2$
To check:
$\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\Rightarrow \frac{3\times 2-2}{4}-\frac{2\times 2+3}{3}=\frac{2}{3}-2$
$\Rightarrow \frac{6-2}{4}-\frac{4+3}{3}=\frac{2-6}{3}\Rightarrow \frac{4}{4}-\frac{7}{3}=\frac{-4}{3}$
$\Rightarrow \frac{1}{1}-\frac{7}{3}=\frac{-4}{3}\Rightarrow \frac{4}{4}-\frac{7}{3}=\frac{-4}{3}$
$\Rightarrow \frac{-4}{3}=\frac{-4}{3}$
$\Rightarrow L.H.S.=R.H.S$
Therefore,it is correct