Solve for \(a\) in \(a+b=7\).
Solve for \(a\).
\[a+b=7\]
Subtract \(b\) from both sides.
\[a=7-b\]
\[a=7-b\]
Substitute \(a=7-b\) into \(ab=12\).
Start with the original equation.
\[ab=12\]
Let \(a=7-b\).
\[(7-b)b=12\]
Simplify.
\[b(7-b)=12\]
\[b(7-b)=12\]
Solve for \(b\) in \(b(7-b)=12\).
Solve for \(b\).
\[b(7-b)=12\]
Expand.
\[7b-{b}^{2}=12\]
Move all terms to one side.
\[7b-{b}^{2}-12=0\]
Multiply both sides by \(-1\).
\[{b}^{2}-7b+12=0\]
Factor \({b}^{2}-7b+12\).
Ask: Which two numbers add up to \(-7\) and multiply to \(12\)?
Rewrite the expression using the above.
\[(b-4)(b-3)\]
\[(b-4)(b-3)=0\]
Solve for \(b\).
Ask: When will \((b-4)(b-3)\) equal zero?
When \(b-4=0\) or \(b-3=0\)
Solve each of the 2 equations above.
\[b=4,3\]
\[b=4,3\]
\[b=4,3\]
Substitute \(b=4,3\) into \(a=7-b\).
Start with the original equation.
\[a=7-b\]
Let \(b=4,3\).
\[a=7-4,7-3\]
Simplify.
\[a=3,4\]
\[a=3,4\]
Therefore,
\[\begin{aligned}&a=3,4\\&b=4,3\end{aligned}\]
a=3,4;b=4,3
