Graph each inequality separately. The overlapping region will be the solution of the system.
Graph:
Area: 25 square units
Practice makes perfect
To solve the given system by graphing, we should first draw each inequality separately. Then we will combine the graphs. The overlapping region will be the solution set. Let's start!
Inequality I
To determine the boundary line of the first inequality, we need to exchange the inequality symbol for an equals sign.
Inequality:& y ≤ 5
Boundary Line:& y=5
This boundary line is a horizontal line. The inequality y ≤ 5 describes all values of y that are less than or equal to 5. This means that every coordinate pair with an y-value that is less than or equal to 5 needs to be included in the shaded region. Notice that the inequality is non-strict, so the boundary line will be solid.
Inequality II
Now that we have completed the first inequality, let's determine the boundary line of the second inequality. We will follow the same process once more.
Inequality:& y > |x+3|
Boundary Line:& y = |x+3|
The graph of this boundary line is the graph of the parent function y=|x| translated left 3 units. The boundary line will be dashed because the inequality is strict.
Next, we need to decide which side of the boundary line we should shade. We can do this by testing a point that does not lie on the boundary line. If the point satisfies the inequality, it lies in the solution set. If not, we will shade the other region. Let's use (0,0).
Because (0,0) created a false statement, we will shade the region that does not contain this point.
Combining the Inequality Graphs
In drawing the inequality graphs on the same coordinate plane, we are able to see the overlapping section. This is the solution set of the system.
We can now view only the solution set by removing the shaded regions that are not overlapping.
Finding the Area
Finally, we will find the area of the intersection of the given inequalities. To do so, note that the shaded region represents a triangle. To calculate the triangle's area, we need to find its base b and its height h.
A = 1/2 b h
We can read the values of b and h from the graph.
Let's now substitute b= 10 and the height h= 5 into the formula for the area of a triangle.
