Let x represent the number of hours that Louie works as a busboy and y represent the number of hours he works as a clerk. Begin by writing a system of inequalities that represents the situation.
Practice makes perfect
Let x represent the number of hours that Louie works as a busboy and y represent the number of hours he works as a clerk.
Job
Pay
Busboy
$6.50
Clerk
$8.00
Knowing that he can work no more than 25, we can say that the sum of the working hours is less than or equal to 25.
x+ y ≤ 25
For the inequality that represents the pay, we will make an organized table.
x+y ≤ 25 & (I) 6.5x+8y≥ 150 & (II)
To graph the system, we will start from the first inequality. Let's determine its boundary line.
&Inequality I &&Boundary Line I
&x+y ≤ 25 &&x+y = 25
The equation of the boundary line is in standard form. In this case, it would be a better way to graph it by finding its intercepts. We will substitute y= 0 for the x-intercept and x= 0 for the y-intercept.
x+y=25
Operation
x-intercept
y-intercept
Substitution
x+ 0=25
0+y=25
Calculation
x=25
y=25
Point
(25,0)
(0,25)
Now we can plot the intercepts and connect them with a line segment. Notice that number of working hours cannot be negative, so it will be bound by the axes. The boundary line will also be solid because of the non-strict inequality.
Next, we should decide which region we will shade. To do so we will test the point (0,0).
Because the point satisfied the inequality, we will shade the region that contains the point. Let's do it!
The second inequality can be graphed by proceeding in the same way.
&Inequality II &&Boundary Line II
&6.5x+8y ≥ 150 &&6.5x+8y = 150
Next step is to determine the intercepts.
6.5x+8y = 150
Operation
x-intercept
y-intercept
Substitution
6.5x+8( 0) = 150
6.5( 0)+8y = 150
Calculation
x=23.1
y=18.75
Point
(23.1,0)
(0,18.75)
Now that we know the intercepts, we should decide which region we will shade as our final step.
