Question
Solve the following equations. Check your result in case. $2\text{x}-\frac{1}{3}=\frac{1}{5}-\text{x}$
✓
Answer
$2\text{x}-\frac{1}{3}=\frac{1}{5}-\text{x}$
$\Rightarrow2\text{x}+\text{x}=\frac{1}{5}+\frac{1}{3}$ (By transposing)
$\Rightarrow3\text{x}=\frac{3+5}{15}$
$\Rightarrow3\text{x}=\frac{8}{15}$
$\Rightarrow\text{x}=\frac{8}{15}\times\frac{1}{3}=\frac{8}{45}$
$\text{x}=\frac{8}{45}$ Check: $\text{L.H.S.}=2\text{x}-\frac{1}{3}$
$=2\times\frac{8}{45}-\frac{1}{3}=\frac{16}{45}-\frac{1}{3}$
$=\frac{16-15}{45}=\frac{1}{45}$
$\text{R.H.S}.=\frac{1}{5}-\text{x}=\frac{1}{5}-\frac{8}{45}$
$=\frac{9-8}{45}=\frac{1}{45}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$ Hence $\text{x}=\frac{8}{45}$
$\Rightarrow2\text{x}+\text{x}=\frac{1}{5}+\frac{1}{3}$ (By transposing)
$\Rightarrow3\text{x}=\frac{3+5}{15}$
$\Rightarrow3\text{x}=\frac{8}{15}$
$\Rightarrow\text{x}=\frac{8}{15}\times\frac{1}{3}=\frac{8}{45}$
$\text{x}=\frac{8}{45}$ Check: $\text{L.H.S.}=2\text{x}-\frac{1}{3}$
$=2\times\frac{8}{45}-\frac{1}{3}=\frac{16}{45}-\frac{1}{3}$
$=\frac{16-15}{45}=\frac{1}{45}$
$\text{R.H.S}.=\frac{1}{5}-\text{x}=\frac{1}{5}-\frac{8}{45}$
$=\frac{9-8}{45}=\frac{1}{45}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$ Hence $\text{x}=\frac{8}{45}$
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