Question
Solve the following equations. Check your result in case.
$\frac{\text{y}+7}{3}=1+\frac{3\text{y}-2}{5}$
$\frac{\text{y}+7}{3}=1+\frac{3\text{y}-2}{5}$
✓
Answer
$\frac{\text{y}+7}{3}=1+\frac{3\text{y}-2}{5}$
$\frac{5(\text{y}+7)=15+3(3\text{y}-2)}{15}$ (LCM of 3, 5 = 15)
$5\text{y}+35=15+9\text{y}-6$
$5\text{y}-9\text{y}=15-6-25$ (By transposing)
$-4\text{y}=-26$
$\text{y}=\frac{-26}{-4}=\frac{13}{2}$
$\therefore\text{y}=\frac{13}{2}$
Check:
$\text{L.H.S.}=\frac{\text{y}-7}{3}=\frac{\frac{13}{2}+7}{3}$
$=\frac{13+14}{2\times3}$
$=\frac{27}{6}=\frac{9}{2}$
$\text{R.H.S.}=1+\frac{3\text{y}-2}{5}$
$=1+\frac{3\times\frac{13}{2}-2}{5}$
$=1+\frac{\frac{39}{2}-\frac{2}{1}}{5}$
$=1+\frac{39-4}{2\times5}$
$=1+\frac{35}{10}$
$=1+\frac{7}{2}=\frac{2+7}{2}=\frac{9}{2}$
$\therefore\text{L.H.S. = R.H.S.}$
Hence $\text{y}=\frac{13}{2}$
$\frac{5(\text{y}+7)=15+3(3\text{y}-2)}{15}$ (LCM of 3, 5 = 15)
$5\text{y}+35=15+9\text{y}-6$
$5\text{y}-9\text{y}=15-6-25$ (By transposing)
$-4\text{y}=-26$
$\text{y}=\frac{-26}{-4}=\frac{13}{2}$
$\therefore\text{y}=\frac{13}{2}$
Check:
$\text{L.H.S.}=\frac{\text{y}-7}{3}=\frac{\frac{13}{2}+7}{3}$
$=\frac{13+14}{2\times3}$
$=\frac{27}{6}=\frac{9}{2}$
$\text{R.H.S.}=1+\frac{3\text{y}-2}{5}$
$=1+\frac{3\times\frac{13}{2}-2}{5}$
$=1+\frac{\frac{39}{2}-\frac{2}{1}}{5}$
$=1+\frac{39-4}{2\times5}$
$=1+\frac{35}{10}$
$=1+\frac{7}{2}=\frac{2+7}{2}=\frac{9}{2}$
$\therefore\text{L.H.S. = R.H.S.}$
Hence $\text{y}=\frac{13}{2}$
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