Question

Solve the following equations and verify your answer :

$\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$

Answer 1

$\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}\Rightarrow \frac{y-7+8y}{9y-3-4y}=\frac{2}{3}\Rightarrow \frac{9y-7}{5y-3}=\frac{2}{3}$
By cross multiplication :
$3(9y-7)=2(5y-3)\Rightarrow 27-21=10y-6$
$\Rightarrow 27-10y=-6+21$ (By transposition)
$\Rightarrow 17y=15\Rightarrow y=\frac{15}{17}\therefore y=\frac{15}{17}$
Verification :
$L.H.S.=\frac{y-(7-8y)}{9y-(3+4y)}=\frac{y-7+8y}{9y-3-4y}=\frac{9y-7}{5y-3}=\frac{9\left(\frac{15}{17}\right)-7}{5\left(\frac{15}{17}\right)-3}=\frac{\frac{16}{17}}{\frac{24}{17}}=\frac{16}{17}\times \frac{17}{24}=\frac{2}{3}=R.H.S.$