Completing a Table of Solutions to a Linear Equation

Completing a Table of Solutions to a Linear Equation

In the previous examples, we substituted the \(x\text{- and}\phantom{\rule{0.2em}{0ex}}y\text{-values}\) of a given ordered pair to determine whether or not it was a solution to a linear equation. But how do we find the ordered pairs if they are not given? One way is to choose a value for \(x\) and then solve the equation for \(y.\) Or, choose a value for \(y\) and then solve for \(x.\)

We’ll start by looking at the solutions to the equation \(y=5x-1\) we found in the last example in the previous lesson. We can summarize this information in a table of solutions.

To find a third solution, we’ll let \(x=2\) and solve for \(y.\)

	\(y=5x-1\)
Multiply.	\(y=10-1\)
Simplify.	\(y=9\)

The ordered pair is a solution to \(y=5x-1\). We will add it to the table.

We can find more solutions to the equation by substituting any value of \(x\) or any value of \(y\) and solving the resulting equation to get another ordered pair that is a solution. There are an infinite number of solutions for this equation.

Example

Complete the table to find three solutions to the equation \(y=4x-2\text{:}\)

\(y=4x-2\)
\(x\)	\(y\)	\(\left(x,y\right)\)
\(0\)
\(-1\)
\(2\)

Solution

Substitute \(x=0,x=-1,\) and \(x=2\) into \(y=4x-2.\)

\(y=4x-2\)	\(y=4x-2\)	\(y=4x-2\)
\(y=0-2\)	\(y=-4-2\)	\(y=8-2\)
\(y=-2\)	\(y=-6\)	\(y=6\)
\(\left(0,-2\right)\)	\(\left(-1,-6\right)\)	\(\left(2,6\right)\)

The results are summarized in the table.

\(y=4x-2\)
\(x\)	\(y\)	\(\left(x,y\right)\)
\(0\)	\(-2\)	\(\left(0,-2\right)\)
\(-1\)	\(-6\)	\(\left(-1,-6\right)\)
\(2\)	\(6\)	\(\left(2,6\right)\)