Question

Solve the following equations and verify your answer :

$\frac{1-9y}{19-3y}=\frac{5}{8}$

Answer 1

$\frac{1-9y}{19-3y}=\frac{5}{8}$
By cross multiplication :
$5(19-3y)=8(1-9y)\Rightarrow 95-15y=8-72y\Rightarrow -15y+72y=8-95\Rightarrow 57y=-87\Rightarrow y=\frac{-87}{57}=\frac{-29}{19}\therefore y=\frac{-29}{19}$
Verification :
$L.H.S.=\frac{1-9y}{19-3y}=\frac{1-9\left(\frac{-29}{19}\right)}{19-3\left(\frac{-29}{19}\right)}=\frac{1+\frac{261}{19}}{19+\frac{87}{19}}=\frac{\frac{19+261}{19}}{\frac{361+87}{19}}=\frac{\frac{280}{19}}{\frac{448}{19}}=\frac{280}{19}\times \frac{19}{448}=\frac{280}{448}=\frac{280÷56}{448÷56}=\frac{5}{8}=R.H.S.$