Question
$2\text{y}-\frac12=-\frac13$
✓
Answer
$2\text{y}-\frac12=-\frac{1}{3}$ Adding $\frac12$ to both sides,
we get $\text{2y}-\frac{1}{2}+\frac12=-\frac13+\frac12$
$\text{y}=\frac{-2+3}{6}$
$\text{2y}=\frac{1}{6}$ Dividing both sides by $2$,
we get $\frac{2\text{y}}{2}=\frac{16}{2}$
$\text{y}=\frac{1}{12}$
Verification: Substituting $\text{y}=\frac{1}{12}$ in $L.H.S$. we get
$\text{L.H.S.}=2\frac{1}{12}-\frac12$
$=\frac16-\frac12$
$=\frac{1-3}{6}$
$=\frac{-2}{6}$
$=-\frac{1}{3},$ and R.H.S. $=-\frac13$
$L.H.S. = R.H.S.$
Hence, verified.
we get $\text{2y}-\frac{1}{2}+\frac12=-\frac13+\frac12$
$\text{y}=\frac{-2+3}{6}$
$\text{2y}=\frac{1}{6}$ Dividing both sides by $2$,
we get $\frac{2\text{y}}{2}=\frac{16}{2}$
$\text{y}=\frac{1}{12}$
Verification: Substituting $\text{y}=\frac{1}{12}$ in $L.H.S$. we get
$\text{L.H.S.}=2\frac{1}{12}-\frac12$
$=\frac16-\frac12$
$=\frac{1-3}{6}$
$=\frac{-2}{6}$
$=-\frac{1}{3},$ and R.H.S. $=-\frac13$
$L.H.S. = R.H.S.$
Hence, verified.
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