Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
6. Systems of Linear Inequalities
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Exercise 31 Page 404

Can someone purchase a part of a shirt?

We can buy 1 dress shirt and 5 T-shirts or 3 dress shirts and 2 T-shirts.

Practice makes perfect

We are given information about spending a gift card at a clothing store. Let's start by assigning variables.

  • The number of T-shirts purchased: x
  • The number of dress shirts purchased: y

We know that the cost of a T-shirt is $15 and the cost of a dress shirt is $22. We can spend no more than the amount of the gift card, $100. Let's rewrite this information as an inequality. 15x+22y≤100 Since we want to leave at most $10 of the gift card unspent, we need to spend at least $90. This creates another inequality. 15x+22y≥90The last constraint that we are given is that we need to buy at least one dress shirt. This gives us our final inequality. y≥1 We now have three inequalities that we can use to find our solution set. 15x+22y≤100 & (I) 15x+22y≥90 & (II) y≥1 & (III) To find all the possible combinations of dress shirts and T-shirts we can buy, we will graph these inequalities in a single coordinate plane. In order to graph the inequalities, we will need to replace the inequality signs with equals signs and isolate the y-variable so that our boundary lines are in slope-intercept form. Let's start with the first inequality.

15x+22y=100
â–¼
Solve for y
22y=100-15x
y=100/22-15x/22
y=50/11-15x/22
y=50/11-15/22x
y=-15/22x+50/11

We can apply the same process to the second inequality as well.

15x+22y=90
â–¼
Solve for y
22y=90-15x
y=90/22-15x/22
y=45/11-15x/22
y=45/11-15/22x
y=-15/22x+45/11

We now have the equations for two of the boundary lines. Let's graph them. We will start with the first equation. Since the inequality is non-strict, the line will be solid. Also, since it is impossible to buy a negative number of shirts, we will only consider positive values of x and y.

Next we will use a test point, (0,0), to find which region to shade. If substituting the point into the inequality creates a true statement, we will shade the region that contains the point. Otherwise, we will shade the region that does not contain the point.

15x+22y≤100
15( 0)+22( 0)? ≤100
0≤100 ✓

Since substituting the test point made the inequality true, we will shade the region that contains it.

We will follow the same process to graph the second inequality. We have already found the equation of the boundary line and because the inequality is non-strict, the line will be solid. We will use the same test point to determine which region to shade.

15x+22y≥90
15( 0)+22( 0)? ≥90
0≱90 *

Since substituting the test point made the inequality false, we will shade the region that does not include it.

Our last inequality, y≥1, is going to be a solid horizontal line. The symbol greater than indicates that the solution set will be above the line.

With most systems of linear inequalities, the solution set includes all points that are in the region shaded by all of the inequalities. Since we are looking at the purchase of shirts, we must only include positive whole number points in our solution, since it is impossible to buy a part of a shirt.

This leaves us with only two possible solutions, (5,1) and (2,3). We can purchase 1 dress shirt and 5 T-shirts, or we can purchase 3 dress shirts and 2 T-shirts.