Question
Solve equation by factorization : $\frac{2}{3} x^2-\frac{1}{3} x=1$
✓
Answer
$
\begin{aligned}
& \frac{2}{3} x^2-\frac{1}{3} x=1 \\
& \Rightarrow 2 x^2-x=3 \\
& \Rightarrow 2 x^2-x-3=0 \\
& \Rightarrow 2 x^2-3 x+2 x-3=0 \\
& \Rightarrow x(2 x-3)+1(2 x-3)=0 \\
& \Rightarrow(2 x-3)(x+1)=0
\end{aligned}
$
Either $2 x-3=0$,
then $2 x =3$
$
\Rightarrow x =\frac{3}{2}
$
or
$
x+1=0 \text {, }
$
then $x=-1$
Hence $x=\frac{3}{2},-1$.
\begin{aligned}
& \frac{2}{3} x^2-\frac{1}{3} x=1 \\
& \Rightarrow 2 x^2-x=3 \\
& \Rightarrow 2 x^2-x-3=0 \\
& \Rightarrow 2 x^2-3 x+2 x-3=0 \\
& \Rightarrow x(2 x-3)+1(2 x-3)=0 \\
& \Rightarrow(2 x-3)(x+1)=0
\end{aligned}
$
Either $2 x-3=0$,
then $2 x =3$
$
\Rightarrow x =\frac{3}{2}
$
or
$
x+1=0 \text {, }
$
then $x=-1$
Hence $x=\frac{3}{2},-1$.
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