Investigate this idea to develop a process for ** isolating a variable**.

- Use the equation \(3x-2y=6\).

- Substitute the value \(x=4\) into the equation \(3x-2y=6\). Isolate the variable \(y\) by solving the equation for \(y\).

\[\begin{align} 3x-2y&=6 \\\\ 3(4)-2y&=6 \\\\ 12-2y&=6 \\\\ 12-12-2y&=6-12 \\\\ -2y&=-6 \\\\ \dfrac{-2y}{-2}&=\dfrac{-6}{-2} \\\\ y&=3\end{align}\]

- Substitute the value \(x=-6\) into the equation \(3x-2y=6\). Isolate the variable \(y\) by solving the equation for \(y\).

\[\begin{align}3x-2y&=6 \\\\ 3(-6)-2y&=6 \\\\ -18-2y&=6 \\\\ -18+18-2y&=6+18 \\\\ -2y&=24 \\\\ \dfrac{-2y}{-2}&=\dfrac{24}{-2} \\\\y&=-12\end{align}\]

- Now solve the equation \(3x-2y=6\) for \(y\) without knowing a value for \(x\). Leave your answer in terms of \(x\) which means that “\(x\)” will remain in your final equation.

\[\begin{align} 3x-2y&=6 \\\\ 3x-3x-2y&=6-3x \\\\ -2y&=-3x+6 \\\\ \dfrac{-2y}{-2}&=\dfrac{-3x+6}{-2} \\\\ y&=\dfrac{3}{2}x-3\end{align}\]

- What is the slope of the line \(3x-2y=6\)? The slope = \(\dfrac{3}{2}\)
- What is the \(y\)-intercept of the line \(3x-2y=6\)? The \(y\)-intercept is \((0,-3)\)

- Solve the equation \(4x+2y=24\) for “\(y\)” in terms of “\(x\)” and identify the slope and the \(y\)-intercept.

\[\begin{align} 4x+2y&=24 \\\\ 4x-4x+2y&=-4x+24 \\\\ \dfrac{2y}{2}&=\dfrac{-4x+24}{2} \\\\ y&=-2x+12\end{align}\]

- The slope = \(-2\)
- The \(y\)-intercept is \((0,12)\)

Return to Try This! (Converting between Slope-Intercept and Standard Form Solutions)