Mathematics » Introducing Graphs » Use the Rectangular Coordinate System

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.

\(y=5x-1\)
\(x\)\(y\)\(\left(x,y\right)\)
\(0\)\(-1\)\(\left(0,-1\right)\)
\(1\)\(4\)\(\left(1,4\right)\)
   

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.

\(y=5x-1\)
\(x\)\(y\)\(\left(x,y\right)\)
\(0\)\(-1\)\(\left(0,-1\right)\)
\(1\)\(4\)\(\left(1,4\right)\)
\(2\)\(9\)\(\left(2,9\right)\)

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.

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

Example

Complete the table to find three solutions to the equation \(5x-4y=20\text{:}\)

\(5x-4y=20\)
\(x\)\(y\)\(\left(x,y\right)\)
\(0\)  
 \(0\) 
 \(5\) 

Solution

The figure shows three algebraic substitutions into an equation. The first substitution is x = 0, with 0 shown in blue. The next line is 5 x- 4 y = 20.  The next line is 5 times 0, shown in blue - 4 y = 20.  The next line is 0 - 4 y = 20.  The next line is - 4 y = 20. The next line is y = -5.   The last line is “ordered pair 0, -5”. The second substitution is y = 0, with 0 shown in red. The next line is 5 x- 4 y = 20.  The next line is 5 x - 4 times 0, with 0 shown in red. The next line is 5 x  - 0 = 20.  The next line is 5 x = 20. The next line is x = 4.   The last line is “ordered pair 4, 0”. The third substitution is  y = 5, with 5 shown in red.  The next line is 5 x- 4 y = 20.  The next line is 5 x - 4 times 5, with 5 shown in blue. The next line is 5 x  - 20 = 20.  The next line is 5 x = 40. The next line is x = 8.   The last line is “ordered pair 8, 5”.

The results are summarized in the table.

\(5x-4y=20\)
\(x\)\(y\)\(\left(x,y\right)\)
\(0\)\(-5\)\(\left(0,-5\right)\)
\(4\)\(0\)\(\left(4,0\right)\)
\(8\)\(5\)\(\left(8,5\right)\)

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