Cumulative Standards Review
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c≥ a c+a≤ 75
Graph:
Is it possible? Yes.
Next, we know that the restaurant has a maximum seating capacity of 75 people. Therefore, the total number of children and adults must be less than or equal to 75. c+a≤75 Now we will combine these inequalities to create one system of inequalities. c≥ a & (I) c+a≤75 & (II)
We can tell a lot of information about the boundary lines from the inequalities given in the system.
Let's find each of these key pieces of information for the inequalities in the system.
| Information | Inequality (I) | Inequality (II) |
|---|---|---|
| Given Inequality | c≥a | c+a≤75 |
| Boundary Line Equation | c=a | c+a=75 |
| Solid or Dashed? | ≥ ⇒ Solid | ≤ ⇒ Solid |
| a= mc+ b | a= 1c+ 0 | a= -1c+ 75 |
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 ( 10, 25) would be a good test point. We will substitute this point for c and a 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 | c≥ a | c+a≤ 75 |
| Substitute (10,25) | 10? ≥ 25 | 10+ 25? ≤75 |
| Simplify | 10≱25 | 10≤75 |
| Shaded Region | opposite | same |
For Inequality (I) we will shade the region opposite our test point, or below the boundary line. For Inequality (II) we will shade the region containing the test point, also below the boundary line.
Now, as we have a shaded solution set we can determine whether it is possible for 50 children to escort 10 adults to the restaurant. Let's plot this point on our graph and see if it lies inside the solution set.
Since this point lies inside the solution set, this combination of children and adults is possible.