Question
$\frac{\text{x}}{2}+\frac32=\frac{2\text{x}}{5}-1$
✓
Answer
$\frac{\text{x}}{2}+\frac{3}{2}=\frac{2\text{x}}5-1{}$
Transposing $\frac{2\text{x}}{5}$ to L.H.S. and $\frac32$ to R.H.S., we get
$=\frac{\text{x}}{2}-\frac{2\text{x}}{5}=-1-\frac32$
$=\frac{5\text{x}-4\text{x}}{10}=\frac{-2-3}{2}$
$=\frac{\text{x}}{10}=\frac{-5}{2}$
Multiplying both sides by 10, we get
$=\frac{\text{x}}{10}\times10=\frac{-5}{2}\times10$
$=\text{x}=-25$
Verification:
Substituting x = -25 in the given equation, we get
$\frac{-25}{2}+\frac32=\frac{2(-25)}{5}-1$
$\frac{-22}{2}=-10-1$
$-11=-11$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.
Transposing $\frac{2\text{x}}{5}$ to L.H.S. and $\frac32$ to R.H.S., we get
$=\frac{\text{x}}{2}-\frac{2\text{x}}{5}=-1-\frac32$
$=\frac{5\text{x}-4\text{x}}{10}=\frac{-2-3}{2}$
$=\frac{\text{x}}{10}=\frac{-5}{2}$
Multiplying both sides by 10, we get
$=\frac{\text{x}}{10}\times10=\frac{-5}{2}\times10$
$=\text{x}=-25$
Verification:
Substituting x = -25 in the given equation, we get
$\frac{-25}{2}+\frac32=\frac{2(-25)}{5}-1$
$\frac{-22}{2}=-10-1$
$-11=-11$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.
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