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.
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.
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