LINR 1 | Lesson 4 | Try This! (Converting between Slope-Intercept and Standard Form ) Solutions

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

  1. 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 \((0,-3)\)

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

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

Go to Practice (Writing Equations of Lines)