to graph a straight inequality in 2 variables (say, x and y ), very first get y alone on one side. Then think about the connected equation acquired by changing the inequality authorize to an equality sign. The graph the this equation is a line.

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

Finally, pick one allude that is no on either line ( ( 0 , 0 ) is generally the easiest) and also decide even if it is these coordinates satisfy the inequality or not. If lock do, shade the half-plane containing that point. If they don"t, the shade the other half-plane.

Graph every of the inequalities in the system in a comparable way. The systems of the system of inequalities is the intersection region of every the solutions in the system.

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** example 1: **

resolve the mechanism of inequalities by graphing:

y ≤ x − 2 y > − 3 x + 5

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

due to the fact that the inequality is ≤ , no a strictly one, the border heat is solid.

Graph the right 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 no contain the allude ( 0 , 0 ) . The shade the lower half of the line.

Similarly, draw a dashed line because that the related equation that the 2nd inequality y > − 3 x + 5 which has actually a strict inequality. The suggest ( 0 , 0 ) walk not meet the inequality, so the shade the fifty percent that does not contain the point ( 0 , 0 ) .

The systems of the mechanism of inequalities is the intersection region of the remedies of the two inequalities.

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** example 2: **

resolve the system of inequalities through graphing:

2 x + 3 y ≥ 12 8 x − 4 y > 1 x 4

Rewrite the very first two inequalities through y alone ~ above 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 associated equation is y = − 2 3 x + 4 .

due to the fact that the inequality is ≥ , not a strictly one, the border heat is solid.

Graph the straight line.

take into consideration a point that is no on the heat - say, ( 0 , 0 ) - and substitute in the inequality.

0 ≥ − 2 3 ( 0 ) + 4 0 ≥ 4

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

Similarly, attract a dashed line of connected equation the the 2nd inequality y 2 x − 1 4 which has actually a strict inequality. The suggest ( 0 , 0 ) walk not accomplish the inequality, so shade the fifty percent that does no contain the suggest ( 0 , 0 ) .

draw a dashed vertical line x = 4 i m sorry is the related equation that the third inequality.

Here suggest ( 0 , 0 ) satisfies the inequality, so the shade the fifty percent that includes the point.

The equipment of the system of inequalities is the intersection region of the options of the 3 inequalities.