Exercise 5 Page 451

Let's assume x represents the number of packs of printer paper and y represents the number packs of graph paper. Now, we will write an inequality for the number of packs and the cost of packs. Remember that the symbols for at least and at most are ≥ and ≤, respectively.

The Number of Packs		The Cost of Packs
Verbal Expression	Algebraic Expression	Verbal Expression	Algebraic Expression
Number of packs of printer paper	x	Cost of x packs of printer paper	$2* x
Number of packs of graph paper	y	Cost of y packs of graph paper	$3* y
Add the number of packs of printer and graph paper	x+ y	Add the costs of packs of printer and graph paper	$2* x+$3* y
Setup the inequality	x+ y≥6	Setup the inequality	$2* x+$3* y≤$27

Thus, we can form our system of inequalities as the following. x+y≥ 6 & (I) 2x+3y≤ 27 & (II) Now, we will graph the system. Let's start with Inequality I.

Inequality I

In order to graph an inequality, we need a boundary line. The boundary line can be written by replacing the inequality symbol with an equals sign in the inequality as the following. x+y≥6 ⇒ x+y=6

Since the inequality is non-strict the boundary line will be solid. Let's draw it!

Next, we need to decide which region we should shade. To do that, we test a point. If the point satisfies the inequality we shade the region that contains the point. Otherwise, we shade the region that does not contain the point. Let's test (0,0).

x+y≥6

SubstituteII

x= 0, y= 0

0+ 0? ≥6

AddTerms

Add terms

0≱6

Since the point does not satisfy the inequality, we will shade the region that does not contain the point.

Adding Inequality II

Let's determine the boundary line of Inequality II, as we did for Inequality I. 2x+3y≤ 27 ⇒ 2x+3y=27 The inequality is non-strict. Therefore, its boundary line will be solid. Now, let's test the point (0,0).

2x+3y≤ 27

SubstituteII

x= 0, y= 0

2* 0+3* 0? ≤ 27

ZeroPropMult

Zero Property of Multiplication

0+0? ≤ 27

AddTerms

Add terms

0≤27

Since the point satisfies the inequality we will shade region that contains the point.

Final Solution

The final solution set for the system is the overlapping section on the previous graph.

Finally, let's plot the point (8,4) on the graph in order to see whether it is a solution for the system.

Since the point is not in the shaded region, it is not a solution for the system.