Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
1. Rate of Change and Slope
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Exercise 61 Page 300

Practice makes perfect
a We have been asked to represent the solution as an inequality. To do this, we will first need to find the perimeter of the rectangle.
Perimeter_(Rectangle)=2(l+w)
Perimeter_(Rectangle)=2( x+2+ 6)
Now, that we know the perimeter, we can create an inequality. The perimeter must be less than 30 inches and greater than 20 inches. 30>2(x+2+6)>20
b To represent the inequality graphically, we will need to solve it first. We can do this by breaking the inequality into two parts and isolating the variable in each.
30 > 2(x+2+6) and 2(x+2+6) > 20 Now, we can solve. Let's start with 30 > 2(x+2+6).
30 > 2(x+2+6)
â–Ľ
Isolate x
30 > 2(x+8)
30 > 2x+16
14 > 2x
7>x
x<7
Now, let's take a look at the second part.
2(x+2+6) > 20
â–Ľ
Isolate x
2(x+8) > 20
2x+16 > 20
2x>4
x>2
We found that x can be less than 7 and greater than 2. Because these inequalities are strict, we need open circles at 2 and 7 on the number line and we can shade between the circles.
c Now, that we have found the possible values for x, we can represent the possible perimeters of the triangle on a number line. Let's start by finding the perimeter of the triangle.
Perimeter_(Triangle)=s_1+s_2+s_3
Perimeter_(Triangle)=6+(x+4)+2x
Perimeter_(Triangle)=3x+10
In Part B, we found that the value of x is greater than 2 and less than 7. 2
Perimeter_(Triangle)=3x+10
Perimeter_(Triangle)=3* 2+10
Perimeter_(Triangle)=6+10
Perimeter_(Triangle)=16
If x=2, the perimeter would be 16 inches however, we know that x>2. This means that the perimeter of the triangle will be greater than 16. Now, we can check for when x=7.
Perimeter_(Triangle)=3x+10
Perimeter_(Triangle)=3* 7+10
Perimeter_(Triangle)=21+10
Perimeter_(Triangle)=31
If x=7, the perimeter would be 31 inches. However, we know that x<7. Therefore, the perimeter of the triangle will be less than 31. Since x can be any value between 2 and 7, the perimeter of the triangle can be any value between 16 and 31. 16 < Perimeter_(Triangle) < 31 These inequalities are also strict so, in the graph, we place open circles on 16 and 31. Then, we will shade between them.