This is a graph of a linear inequality:
The inequality y ≤ x + 2
We can see the y = x + 2 line, and the shaded area is where y is less than or equal to x + 2
Linear Inequality
A Linear Inequality is like a Linear Equation (such as y = 2x+1) ...
... but it will have an Inequality like <, >, ≤, or ≥ instead of an =.
How to Graph a Linear Inequality
Graph the "equals" line, then shade in the correct area.
Follow these steps:
- Rearrange the equation so "y" is on the left and everything else on the right.
- Plot the "y=" line (make it a solid line for y≤ or y≥, and a dashed line for y< or y>)
- Shade above the line for a "greater than" (y> or y≥)
or below the line for a "less than" (y< or y≤).
Let us try some examples:
Example: y ≤ 2x−1
1. The inequality already has "y" on the left and everything else on the right, so no need to rearrange.
2. Plot y = 2x−1 (as a solid line because y≤ includes equal to):
3. Shade the area below (because y is less than or equal to):
Example: 2y − x ≤ 6
1. We will need to rearrange this one so "y" is on its own on the left:
Start with: 2y − x ≤ 6
Add x to both sides: 2y ≤ x + 6
Divide all by 2: y ≤ x/2 + 3
2. Now plot y = x/2 + 3 (as a solid line because y≤ includes equal to):
3. Shade the area below (because y is less than or equal to):
Example: y/2 + 2 > x
1. We will need to rearrange this one so "y" is on its own on the left:
Start with: y/2 + 2 > x
Subtract 2 from both sides: y/2 > x − 2
Multiply all by 2: y > 2x − 4
2. Now plot y = 2x − 4 (as a dashed line because y> does not include equals to):
3. Shade the area above (because y is greater than):
The dashed line shows that the inequality does not include the line y = 2x−4.
Two Special Cases
We can also have a horizontal, or vertical, line:
7426, 7427, 7428, 7429, 7430, 7431, 7432, 7433, 7434, 7435
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.