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.




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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.

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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.

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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 ) .

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

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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.

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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 ) .

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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.

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