Solve for \(x\) in \(2x-\frac{3}{y}=9\).
Solve for \(x\).
\[2x-\frac{3}{y}=9\]
Add \(\frac{3}{y}\) to both sides.
\[2x=9+\frac{3}{y}\]
Divide both sides by \(2\).
\[x=\frac{9+\frac{3}{y}}{2}\]
Factor out the common term \(3\).
\[x=\frac{3(3+\frac{1}{y})}{2}\]
\[x=\frac{3(3+\frac{1}{y})}{2}\]
Substitute \(x=\frac{3(3+\frac{1}{y})}{2}\) into \(3x+7y=2(yne\times 0)\).
Start with the original equation.
\[3x+7y=2(yne\times 0)\]
Let \(x=\frac{3(3+\frac{1}{y})}{2}\).
\[3\times \frac{3(3+\frac{1}{y})}{2}+7y=2yne\times 0\]
Simplify.
\[\frac{9}{2}(3+\frac{1}{y})+7y=0\]
\[\frac{9}{2}(3+\frac{1}{y})+7y=0\]
Solve for \(y\) in \(\frac{9}{2}(3+\frac{1}{y})+7y=0\).
Solve for \(y\).
\[\frac{9}{2}(3+\frac{1}{y})+7y=0\]
Multiply both sides by \(2\).
\[9(3+\frac{1}{y})+14y=0\]
Expand.
\[27+\frac{9}{y}+14y=0\]
Multiply both sides by \(y\).
\[27y+9+14{y}^{2}=0\]
Split the second term in \(27y+9+14{y}^{2}\) into two terms.
Multiply the coefficient of the first term by the constant term.
\[14\times 9=126\]
Ask: Which two numbers add up to \(27\) and multiply to \(126\)?
Split \(27y\) as the sum of \(21y\) and \(6y\).
\[14{y}^{2}+21y+6y+9\]
\[14{y}^{2}+21y+6y+9=0\]
Factor out common terms in the first two terms, then in the last two terms.
\[7y(2y+3)+3(2y+3)=0\]
Factor out the common term \(2y+3\).
\[(2y+3)(7y+3)=0\]
Solve for \(y\).
Ask: When will \((2y+3)(7y+3)\) equal zero?
When \(2y+3=0\) or \(7y+3=0\)
Solve each of the 2 equations above.
\[y=-\frac{3}{2},-\frac{3}{7}\]
\[y=-\frac{3}{2},-\frac{3}{7}\]
\[y=-\frac{3}{2},-\frac{3}{7}\]
Substitute \(y=-\frac{3}{2},-\frac{3}{7}\) into \(x=\frac{3(3+\frac{1}{y})}{2}\).
Start with the original equation.
\[x=\frac{3(3+\frac{1}{y})}{2}\]
Let \(y=-\frac{3}{2},-\frac{3}{7}\).
\[x=\frac{3(3+\frac{1}{-\frac{3}{2}})}{2},\frac{3(3+\frac{1}{-\frac{3}{7}})}{2}\]
Simplify.
\[x=\frac{7}{2},1\]
\[x=\frac{7}{2},1\]
Therefore,
\[\begin{aligned}&x=\frac{7}{2},1\\&y=-\frac{3}{2},-\frac{3}{7}\end{aligned}\]
x=7/2,1;y=-3/2,-3/7
