Video transcript
Is two comma five a solution of this system? And we have a system of inequalities right over here. We have Y is greater than or equal to 2x plus 1 and X is greater than 1. In order for two comma five to be a solution of this system, it just has to satisfy both inequalities. So, lets just try it out. So when X is equal to two and Y is equal to five, it has to satify both of these. So lets try it with the first one. So if we assume X is two and Y is five, we would get an inequality that says that five is greater than or equal to two times two plus one. X is two; Y is five. This gives us five is greater than or equal to two times two is four plus one is five. Y is greater than or equal to five. That's true! Five is equal to five. So that equal part of the greater than or equal saves us. That definitely satisfies the first inequality. Lets see the second one. X needs to be greater than one. So in two comma five, X is two. So two is greater than one. So it actually satisfies both of these inequalities. So two comma five is a solution for this system.
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.