Question
Solve the following equation by factorization
$3(y^2 – 6) = y(y + 7) – 3$
$3(y^2 – 6) = y(y + 7) – 3$
✓
Answer
$
\begin{aligned}
& 3\left(y^2-6\right)=y(y+7)-3 \\
& \Rightarrow 3\left(y^2-6\right)=y^2+7 y-3 \\
& \Rightarrow 3 y^2-18=y^2+7 y-3 \\
& \Rightarrow 3 y^2-y^2-7 y-18+3=0 \\
& \Rightarrow 2 y^2-7 y-15=0 \\
& \Rightarrow 2 y^2-10 y+3 y-15=0 \\
& 2 y(y-5)+3(y-5)=0 \\
& \Rightarrow(y-5)(2 y+3)=0
\end{aligned}
$
Either y - $5=0$,
then $y=5$
or
$
2 y+3=0 \text {, }
$
then $2 y=-3$
$
\Rightarrow y =\frac{-3}{2}
$
Hence $y=\frac{-3}{2}, 5$.
\begin{aligned}
& 3\left(y^2-6\right)=y(y+7)-3 \\
& \Rightarrow 3\left(y^2-6\right)=y^2+7 y-3 \\
& \Rightarrow 3 y^2-18=y^2+7 y-3 \\
& \Rightarrow 3 y^2-y^2-7 y-18+3=0 \\
& \Rightarrow 2 y^2-7 y-15=0 \\
& \Rightarrow 2 y^2-10 y+3 y-15=0 \\
& 2 y(y-5)+3(y-5)=0 \\
& \Rightarrow(y-5)(2 y+3)=0
\end{aligned}
$
Either y - $5=0$,
then $y=5$
or
$
2 y+3=0 \text {, }
$
then $2 y=-3$
$
\Rightarrow y =\frac{-3}{2}
$
Hence $y=\frac{-3}{2}, 5$.
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