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 36 Page 404

Practice makes perfect
a We are given information about a jeweler creating a ring. Let's start by assigning variables.
  • Grams of used:
  • Grams of used:

We are also given three different constraints for the mass and cost of the ring. We can create inequalities from that information.

The mass must be more than grams but less than grams There must be at least grams of used The total cost must be less than
Now that we have the inequalities written we need to graph them. Since one of the inequalities is a compound inequality, we will break that into two simple inequalities. We now have the following system.
Let's start by creating the boundary line equations by isolating the variable for each of the equations.
We now have the equations for all of the boundary lines. Let's graph them. We will start with the first equation. Since the inequality is less than the line will be dashed. When graphing we must keep in mind that neither nor can have negative values.
Diagram showing the graph of y=-x+10
Now we will use a test point, to find out how to shade.
Since the test point made the inequality false we will shade up and away from the point.
Diagram showing the solutions to the inequality y greater than -x+10

We will follow the same process to graph the other inequalities. We have already found the boundary lines. We will use the same test point to determine where to shade.

Inequality Solid or Dashed Line True or False
Dashed True
Solid False
Dashed True

Now we can graph all of the inequalities and shade them appropriately.

Diagram showing the solutions to each of the four inequalities overlapping each other

The solution region is where all of the shading overlaps.

Diagram showing the solution set to the system of inequalities


b In order to find the mass and cost of one of the solutions, we need to choose a point in the shaded region.
Diagram showing the solution set to the system of inequalities and the point (10,3).
We will use the point Note that this is just one point out of many that can be used. We need to use the expressions for the cost and the mass to find the correct values.
We can substitute the values into the first expression to find the mass.
Now, we can substitute the values into the second expression to find the cost.
Therefore, if grams of and grams of are used, the ring will have a mass of grams and cost