Question
Solve the following equation and verify your answer: $\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}=10$
✓
Answer
$\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}=10$
$\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}=\frac{10}{1}$
$\Rightarrow\frac{-20\text{x}}{1-3\text{x}}$
$=\frac{10}{1}$
By cross multiplication, $-20\text{x}=10(1-3\text{x})$
$\Rightarrow-20\text{x}-10=30\text{x}$
$\Rightarrow20\text{x}+30\text{x}=10$
$\Rightarrow10\text{x}=10$
$\Rightarrow\text{x}=\frac{10}{10}=1$
$=1$
$\therefore\text{x}=1$
Verification: $\text{L.H.S}=\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}$
$=\frac{15(2-1)-5(1+6)}{1-3\times1}$
$\frac{15\times1-5\times7}{1-3}-\frac{15-35}{-2}$
$=\frac{-20}{-2}=10$
$=\text{R.H.S}$
$\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}=\frac{10}{1}$
$\Rightarrow\frac{-20\text{x}}{1-3\text{x}}$
$=\frac{10}{1}$
By cross multiplication, $-20\text{x}=10(1-3\text{x})$
$\Rightarrow-20\text{x}-10=30\text{x}$
$\Rightarrow20\text{x}+30\text{x}=10$
$\Rightarrow10\text{x}=10$
$\Rightarrow\text{x}=\frac{10}{10}=1$
$=1$
$\therefore\text{x}=1$
Verification: $\text{L.H.S}=\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}$
$=\frac{15(2-1)-5(1+6)}{1-3\times1}$
$\frac{15\times1-5\times7}{1-3}-\frac{15-35}{-2}$
$=\frac{-20}{-2}=10$
$=\text{R.H.S}$
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