Question
Solve the following equation and verify the answer:
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$
✓
Answer
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$
Multiplying each term by $10$, the $L.C.M$. of $5$ and $2,$
we get $\frac{\text{2x}}{5}\times10-\frac{3}{2}\times10$
$=\frac{\text{x}}{2}\times10+1\times10$
$\Rightarrow 4x - 15 = 5x + 10$
$\Rightarrow 4x - 5x = 10 + 15$
(transposing $5x$ to $L.H.S.)$
$\Rightarrow -x = 25$
$\Rightarrow x = -25$
(multiplying both sides by$ -1)$
So, $x = -25$ is a solution of the given equation.
Check: Substituting $x = -25$ in the given equation, we get
$\text{L.H.S.}=\frac{2\times(-25)}{5}-\frac{3}{2}=-10-\frac{3}{2}$
$=\frac{-20-3}{2}=\frac{-23}{2}$
$\text{R.H.S.}=\frac{-25}{2}+1=\frac{-25+2}{2}=\frac{-23}{2}$
$\therefore$ When$ x = -25$, we have
$L.H.S. = R.H.S.$
Multiplying each term by $10$, the $L.C.M$. of $5$ and $2,$
we get $\frac{\text{2x}}{5}\times10-\frac{3}{2}\times10$
$=\frac{\text{x}}{2}\times10+1\times10$
$\Rightarrow 4x - 15 = 5x + 10$
$\Rightarrow 4x - 5x = 10 + 15$
(transposing $5x$ to $L.H.S.)$
$\Rightarrow -x = 25$
$\Rightarrow x = -25$
(multiplying both sides by$ -1)$
So, $x = -25$ is a solution of the given equation.
Check: Substituting $x = -25$ in the given equation, we get
$\text{L.H.S.}=\frac{2\times(-25)}{5}-\frac{3}{2}=-10-\frac{3}{2}$
$=\frac{-20-3}{2}=\frac{-23}{2}$
$\text{R.H.S.}=\frac{-25}{2}+1=\frac{-25+2}{2}=\frac{-23}{2}$
$\therefore$ When$ x = -25$, we have
$L.H.S. = R.H.S.$
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