Question
Solve:
$\frac{\text{2x}+7}{5}-\frac{\text{3x}+11}{2}=\frac{2\text{x}+8}{3}-5$
$\frac{\text{2x}+7}{5}-\frac{\text{3x}+11}{2}=\frac{2\text{x}+8}{3}-5$
✓
Answer
$\frac{\text{2x}+7}{5}-\frac{\text{3x}+11}{2}=\frac{2\text{x}+8}{3}-5$
$\Rightarrow\frac{2\text{x} + 7}{5} - \frac{3\text{x} + 11}{2} = \frac{2\text{x} + 8 - 15}{3} $ (L.C.M of 3 and 1 is 3)
$\Rightarrow30\Big(\frac{2\text{x} + 7}{5}\Big) -30\Big( \frac{3\text{x} + 11}{2}\Big) =30\Big( \frac{2\text{x} + 8 - 15}{3} \Big)$
(multiplying throughout by 30, which is the L.C.M. of 5, 2 and 3)
$\Rightarrow6 (2\text{x} + 7) - 15 (3\text{x} + 11) = 10 (2\text{x} + 8 - 15) $
$\Rightarrow12\text{x} + 42 -45\text{x} - 165 = 20\text{x} - 70$
$-33\text{x}-123=20\text{x}$
$\Rightarrow\frac{2\text{x} + 7}{5} - \frac{3\text{x} + 11}{2} = \frac{2\text{x} + 8 - 15}{3} $ (L.C.M of 3 and 1 is 3)
$\Rightarrow30\Big(\frac{2\text{x} + 7}{5}\Big) -30\Big( \frac{3\text{x} + 11}{2}\Big) =30\Big( \frac{2\text{x} + 8 - 15}{3} \Big)$
(multiplying throughout by 30, which is the L.C.M. of 5, 2 and 3)
$\Rightarrow6 (2\text{x} + 7) - 15 (3\text{x} + 11) = 10 (2\text{x} + 8 - 15) $
$\Rightarrow12\text{x} + 42 -45\text{x} - 165 = 20\text{x} - 70$
$-33\text{x}-123=20\text{x}$
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