Solve each system of inequalities by graphing calculator

To graph a linear inequality in two variables (say, x and y ), first get y alone on one side. Then consider the related equation obtained by changing the inequality sign to an equality sign. The graph of this equation is a line.

If the inequality is strict ( < or > ), graph a dashed line. If the inequality is not strict ( ≤ or ≥ ), graph a solid line.

Finally, pick one point that is not on either line ( ( 0 , 0 ) is usually the easiest) and decide whether these coordinates satisfy the inequality or not. If they do, shade the half-plane containing that point. If they don't, shade the other half-plane.

Graph each of the inequalities in the system in a similar way. The solution of the system of inequalities is the intersection region of all the solutions in the system.

Example 1:

Solve the system of inequalities by graphing:

y ≤ x − 2 y > − 3 x + 5

First, graph the inequality y ≤ x − 2 . The related equation is y = x − 2 .

Since the inequality is ≤ , not a strict one, the border line is solid.

Graph the straight line.

Consider a point that is not on the line - say, ( 0 , 0 ) - and substitute in the inequality y ≤ x − 2 .

0 ≤ 0 − 2 0 ≤ − 2

This is false. So, the solution does not contain the point ( 0 , 0 ) . Shade the lower half of the line.

Similarly, draw a dashed line for the related equation of the second inequality y > − 3 x + 5 which has a strict inequality. The point ( 0 , 0 ) does not satisfy the inequality, so shade the half that does not contain the point ( 0 , 0 ) .

The solution of the system of inequalities is the intersection region of the solutions of the two inequalities.

Example 2:

Solve the system of inequalities by graphing:

2 x + 3 y ≥ 12 8 x − 4 y > 1 x < 4

Rewrite the first two inequalities with y alone on one side.

3 y ≥ − 2 x + 12 y ≥ − 2 3 x + 4 − 4 y > − 8 x + 1 y < 2 x − 1 4

Now, graph the inequality y ≥ − 2 3 x + 4 . The related equation is y = − 2 3 x + 4 .

Since the inequality is ≥ , not a strict one, the border line is solid.

Graph the straight line.

Consider a point that is not on the line - say, ( 0 , 0 ) - and substitute in the inequality.

0 ≥ − 2 3 ( 0 ) + 4 0 ≥ 4

This is false. So, the solution does not contain the point ( 0 , 0 ) . Shade upper half of the line.

Similarly, draw a dashed line of related equation of the second inequality y < 2 x − 1 4 which has a strict inequality. The point ( 0 , 0 ) does not satisfy the inequality, so shade the half that does not contain the point ( 0 , 0 ) .

Draw a dashed vertical line x = 4 which is the related equation of the third inequality.

Here point ( 0 , 0 ) satisfies the inequality, so shade the half that contains the point.

The solution of the system of inequalities is the intersection region of the solutions of the three inequalities.

We first need to review the symbols for inequalities:

  • The symbol < means less than.
  • The symbol > means greater than.
  • The symbol \(\leq\) means less than or equal to. Usually this is written as <= on computers because it is easier to type.
  • The symbol \(\geq\) means greater than or equal to. Sometimes this is written as >= on computers because it is easier to type.

There are endless solutions for inequalities. In light of this fact, it may be easiest to find a solution set for inequalities by solving the system graphically.

How To Solve Systems of Inequalities Graphically

1) Write the inequality in slope-intercept form or in the form \(y = mx + b\).

For example, if asked to solve \(x + y \leq 10\), we first re-write as \(y \leq -x + 10\).

2) Temporarily exchange the given inequality symbol (in this case \(\leq\)) for just equal symbol. In doing so, you can treat the inequality like an equation. BUT DO NOT forget to replace the equal symbol with the original inequality symbol at the END of the problem!

So, \(y \leq -x + 10\) becomes \(y = -x + 10\) for the moment.

3) Graph the line found in step 2. This will form the "boundary" of the inequality -- on one side of the line the condition will be true, on the other side it will not. Review how to graph a line here.

4) Revisit the inequality we found before as \(y \leq -x + 10\). Notice that it is true when y is less than or equal to. In step 3 we plotted the line (the equal-to case), so now we need to account for the less-than case. Since y is less than a particular value on the low-side of the axis, we will shade the region below the line to indicate that the inequality is true for all points below the line:

5) Verify. Plug in a point not on the line, like (0,0). Verify that the inequality holds. In this case, that means \(0 \leq -0+10\), which is clearly true. We have shaded the correct side of the line.

Example:

Find all values of x and y that satisfy: \(y \geq \frac{-3}{2}x + 6\).

Notice that this inequality is already in the slope-intercept form. I will replace the given inequality symbol for the equal symbol to plot the line.

\(y \geq \frac{-3}{2}x + 6\) becomes \(y = \frac{-3}{2}x + 6\). Now plot that line as shown:

Since this is a case where the inequality is true for y values greater than or equal to something, we have shaded the area above the line. All points on or ABOVE this graph line will satisfy our inequality. Again, select any point above the graph line to make sure that it will satisfy or reveal a TRUE statement in terms of the original inequality. For example, (5,3). Plug that in and we have \(3 \geq \frac{-3}{2}*5+6\). Simplify it to \(3 \geq -1.5\) and we see that the inequality is true at the point (5,3). Since that point was above our line, it should be shaded, which verifies our solution.

Multiple inequalities - a system of inequalities

A system of inequalities has more than one inequality statement that must be satisfied. Graphically, it means we need to do what we just did -- plot the line represented by each inequality -- and then find the region of the graph that is true for BOTH inequalities. For the two examples above, we can combine both graphs and plot the area shared by the two inequalities.

What is the solution set? The solution set for BOTH inequalities will be ANY POINT where BOTH regions are shaded together or where BOTH shaded regions meet.

Originally by Mr. Feliz, © 2005

Can a TI

Your TI-84 Plus calculator was not made to graph inequalities on a number line, but it can be used to accomplish that task. The reason your calculator is able to perform a task that it was not designed for is the Boolean logic your calculator uses to evaluate statements.

How do you solve each system by graphing?

To solve a system of linear equations graphically we graph both equations in the same coordinate system. The solution to the system will be in the point where the two lines intersect. The two lines intersect in (-3, -4) which is the solution to this system of equations.

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