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 \(y\).

\[3x-2y=6\]

\[3(4)-2y=6\]

\[12-2y=6\]

\[12-12-2y=6-12\]

\[-2y=-6\]

\[\frac{-2y}{-2}=\frac{-6}{-2}\]

\[y=3\]


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

\[3x-2y=6\]

\[3(-6)-2y=6\]

\[-18-2y=6\]

\[-18+18-2y=6+18\]

\[-2y=24\]

\[\frac{-2y}{-2}=\frac{24}{-2}\]

\[y=-12\]


  • 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.

\[3x-2y=6\]

\[3x-3x-2y=6-3x\]

\[-2y=-3x+6\]

\[\frac{-2y}{-2}=\frac{-3x+6}{-2}\]

\[y=\frac{3}{2}x-3\]

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

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

\[4x+2y=24\]

\[4x-4x+2y=-4x+24\]

\[\frac{2y}{2}=\frac{-4x+24}{2}\]

\[y=-2x+12\]

 

 

 

Go to Practice (Writing Equations of Lines)