Mathematics » Graphs and Equations » Use the Rectangular Coordinate System

Finding Solutions to a Linear Equation

Finding Solutions to a Linear Equation

To find a solution to a linear equation, you really can pick any number you want to substitute into the equation for \(x\) or \(y.\) But since you’ll need to use that number to solve for the other variable it’s a good idea to choose a number that’s easy to work with.

When the equation is in y-form, with the y by itself on one side of the equation, it is usually easier to choose values of \(x\) and then solve for \(y\).

Example

Find three solutions to the equation \(y=-3x+2\).

Solution

We can substitute any value we want for \(x\) or any value for \(y\). Since the equation is in y-form, it will be easier to substitute in values of \(x\). Let’s pick \(x=0\), \(x=1\), and \(x=-1\).

 
 ...
 ...
Substitute the value into the equation....
Simplify....
Simplify....
Write the ordered pair.(0, 2)(1, −1)(−1, 5)
Check.   
\(\phantom{\rule{0.03em}{0ex}}y=-3x+2\)\(\phantom{\rule{0.7em}{0ex}}y=-3x+2\)\(\phantom{\rule{0.04em}{0ex}}y=-3x+2\)   
\(2\stackrel{?}{=}-3\cdot 0+2\)\(-1\stackrel{?}{=}-3\cdot 1+2\)\(5\stackrel{?}{=}-3\left(-1\right)+2\)   
\(2\stackrel{?}{=}0+2\)\(-1\stackrel{?}{=}-3+2\)\(5\stackrel{?}{=}3+2\)   
\(2=2✓\)\(-1=-1✓\)\(5=5✓\)   

So, \(\left(0,2\right)\), \(\left(1,-1\right)\) and \(\left(-1,5\right)\) are all solutions to \(y=-3x+2\). We show them in the table below.

\(y=-3x+2\)
\(x\)\(y\)\(\left(x,y\right)\)
02\(\left(0,2\right)\)
1\(-1\)\(\left(1,-1\right)\)
\(-1\)5\(\left(-1,5\right)\)

We have seen how using zero as one value of \(x\) makes finding the value of \(y\) easy. When an equation is in standard form, with both the \(x\) and \(y\) on the same side of the equation, it is usually easier to first find one solution when \(x=0\) find a second solution when \(y=0\), and then find a third solution.

Example

Find three solutions to the equation \(3x+2y=6\).

Solution

We can substitute any value we want for \(x\) or any value for \(y\). Since the equation is in standard form, let’s pick first \(x=0\), then \(y=0\), and then find a third point.

 
 ...
 ...
Substitute the value into the equation....
Simplify....
Solve....
 ...
Write the ordered pair.(0, 3)(2, 0)\(\left(1,\frac{3}{2}\right)\)
Check.   
\(3x+2y=6\phantom{\rule{1.3em}{0ex}}\)\(3x+2y=6\phantom{\rule{1.3em}{0ex}}\)\(3x+2y=6\phantom{\rule{1.3em}{0ex}}\)   
\(3\cdot 0+2\cdot 3\stackrel{?}{=}6\phantom{\rule{1.3em}{0ex}}\)\(3\cdot 2+2\cdot 0\stackrel{?}{=}6\phantom{\rule{1.3em}{0ex}}\)\(3\cdot 1+2\cdot \frac{3}{2}\stackrel{?}{=}6\phantom{\rule{1.3em}{0ex}}\)   
\(0+6\stackrel{?}{=}6\phantom{\rule{1.3em}{0ex}}\)\(6+0\stackrel{?}{=}6\phantom{\rule{1.3em}{0ex}}\)\(3+3\stackrel{?}{=}6\phantom{\rule{1.3em}{0ex}}\)   
\(6=6✓\)\(6=6✓\)\(6=6✓\)   

So \(\left(0,3\right)\), \(\left(2,0\right)\), and \(\left(1,\frac{3}{2}\right)\) are all solutions to the equation \(3x+2y=6\). We can list these three solutions in the table below.

\(3x+2y=6\)
\(x\)\(y\)\(\left(x,y\right)\)
03\(\left(0,3\right)\)
20\(\left(2,0\right)\)
1\(\frac{3}{2}\)\(\left(1,\frac{3}{2}\right)\)

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