Question
Solve the following equation and verify your answer: $\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{7}{6}$
✓
Answer
$\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{7}{6}$
$\Rightarrow\frac{2\text{x}-7+5\text{x}}{9\text{x}-3-4\text{x}}=\frac{7}{6}$
$\Rightarrow\frac{7\text{x}-7}{5\text{x}-3}=\frac{7}{6}$
By cross multiplication, $6(7\text{x}-7)=7(5\text{x}-3)$
$\Rightarrow42\text{x}-42=35\text{x}-21$
$\Rightarrow42\text{x}-35\text{x}=-21+42$
$\Rightarrow7\text{x}=21$
$\Rightarrow\text{x}=\frac{21}{7}=3$
$=3$
$\therefore\text{x}=3$
Verification: $\text{L.H.S}=\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{2\times3-(7-5\times3)}{9\times3-(3+4\times3)}$
$=\frac{6-(7-15)}{27-(3+12)}=\frac{6-7+15}{27-3-12}=\frac{14}{12}$
$=\frac{7}{6}$
$=\text{R.H.S}$
$\Rightarrow\frac{2\text{x}-7+5\text{x}}{9\text{x}-3-4\text{x}}=\frac{7}{6}$
$\Rightarrow\frac{7\text{x}-7}{5\text{x}-3}=\frac{7}{6}$
By cross multiplication, $6(7\text{x}-7)=7(5\text{x}-3)$
$\Rightarrow42\text{x}-42=35\text{x}-21$
$\Rightarrow42\text{x}-35\text{x}=-21+42$
$\Rightarrow7\text{x}=21$
$\Rightarrow\text{x}=\frac{21}{7}=3$
$=3$
$\therefore\text{x}=3$
Verification: $\text{L.H.S}=\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{2\times3-(7-5\times3)}{9\times3-(3+4\times3)}$
$=\frac{6-(7-15)}{27-(3+12)}=\frac{6-7+15}{27-3-12}=\frac{14}{12}$
$=\frac{7}{6}$
$=\text{R.H.S}$
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