Question
Solve the following equation and verify your answer: $\frac{1-9\text{y}}{19-3\text{y}}=\frac{5}{8}$
✓
Answer
$\frac{1-9\text{y}}{19-3\text{y}}=\frac{5}{8}$
By cross multiplication:
$5(19-3\text{y})=8(1-9\text{y})$
$\Rightarrow95-15\text{y}=8-72\text{y}$
$\Rightarrow15\text{y}+72\text{y}=8-95$
$\Rightarrow57\text{y}=-87$
$\Rightarrow\text{y}=\frac{-87}{57}=\frac{-29}{19}$
$\therefore\text{y}=\frac{-29}{19}$
Verification:
$\text{L.H.S.}=\frac{1-9\text{y}}{19-3\text{y}}=\frac{1-9\Big(\frac{-29}{19}\Big)}{19-3\Big(\frac{-29}{19}\Big)}=\frac{1+\frac{261}{19}}{91+\frac{87}{19}}$
$=\frac{\frac{19+261}{19}}{\frac{361+87}{19}}=\frac{\frac{280}{19}}{\frac{448}{19}}\times\frac{19}{448}$
$=\frac{280}{448}=\frac{280\div56}{448\div56}=\frac{5}{8}=\text{R.H.S.}$
By cross multiplication:
$5(19-3\text{y})=8(1-9\text{y})$
$\Rightarrow95-15\text{y}=8-72\text{y}$
$\Rightarrow15\text{y}+72\text{y}=8-95$
$\Rightarrow57\text{y}=-87$
$\Rightarrow\text{y}=\frac{-87}{57}=\frac{-29}{19}$
$\therefore\text{y}=\frac{-29}{19}$
Verification:
$\text{L.H.S.}=\frac{1-9\text{y}}{19-3\text{y}}=\frac{1-9\Big(\frac{-29}{19}\Big)}{19-3\Big(\frac{-29}{19}\Big)}=\frac{1+\frac{261}{19}}{91+\frac{87}{19}}$
$=\frac{\frac{19+261}{19}}{\frac{361+87}{19}}=\frac{\frac{280}{19}}{\frac{448}{19}}\times\frac{19}{448}$
$=\frac{280}{448}=\frac{280\div56}{448\div56}=\frac{5}{8}=\text{R.H.S.}$
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