Chapter Test
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Graph each inequality separately. The solution will be the intersection, or overlap, of the shaded regions.
Graphing a single inequality involves two main steps.
Here, we need to do this process for each of the inequalities in the system. y> 12x+4 & (I) 2y≤ x+4 & (II) The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).
Let's find each of these key pieces of information for the inequalities in the system.
| Information | Inequality (I) | Inequality (II) |
|---|---|---|
| Given Inequality | y > 12x+4 | 2y ≤x+4 |
| Boundary Line Equation | y = 12x+4 | 2y =x+4 |
| Solid or Dashed? | > ⇒ Dashed | ≤ ⇒ Solid |
| y= mx+ b | y= 12x+ 4 | y= 12x+ 2 |
Great! With all of this information, we can plot the boundary lines.
Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.
It looks like the point ( 0, 0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region as the test point. Otherwise, we shade the opposite region.
| Information | Inequality (I) | Inequality (II) |
|---|---|---|
| Given Inequality | y> 12x+4 | 2y≤ x+4 |
| Substitute (0,0) | 0? > 12( 0)+4 | 2( 0)? ≤( 0)+4 |
| Simplify | 0≯4 * | 0≤4 ✓ |
| Shaded Region | opposite | same |
For Inequality (I), we will shade the region opposite test point, or above the boundary line. For Inequality (II), however, we will shade the region containing our test point, or below the boundary line.
Now that we have graphed the system, we see that there is no overlapping region. This means that there are no points that are a solution to the system, only points that are solutions to each individual inequality.