Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 11.2
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Exercise 38 Page 538

Practice makes perfect
a We are asked to solve the given inequality and graph the solution set on the number line.
|x|+3 < 5

Let's begin by gathering all constant terms on the right-hand side of the inequality. |x|+3 < 5 ⇔ |x|<2 Now, we can create a compound inequality by removing the absolute value. In this case, the solution set is any number less than 2 away from the midpoint in the positive direction and any number less than 2 away from the midpoint in the negative direction. Absolute Value Inequality:& |x| < 2 Compound Inequality:& - 2< x < 2 The graph of this inequality includes all values from - 2 to 2, not inclusive. We show this by using open circles on the endpoints.

b Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
5(2x+1) ≥ 30
10x+5 ≥ 30
â–Ľ
Simplify
10x ≥ 25
x ≥ 25/10
x ≥ 2.5
This inequality tells us that all values greater than or equal to 2.5 will satisfy the inequality. Below we demonstrate the inequality by graphing the solution set on a number line. Notice that x can be equal to 2.5, which we show with an closed circle on the number line.
c To solve the given equation we should start by isolating the x-term on the left-hand side of the equation.
1/x-5/2=3/2
1/x=8/2
Now, to simplify the equation with fractions on both sides of the equal sign we can use the Cross Product Property.
1/x=8/2
x * 8 = 1 * 2
â–Ľ
Solve for x
8x=2
x=2/8
x=1/4
The solution to the equation is x= 14. Let's represent it on a number line!
d Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
- 5-x >3-x
- 5 ≯ 3 *
The given inequality results in a contradiction. It means that there is no value of x that will satisfy the inequality. Thus, there is no solution. We can represent this graphically on a number line by showing that no values are included in the solution set, so no values are shaded.
e To solve the given equation, let's isolate the square root on one side of the given equation first.
3 sqrt(4-x)+1=13
3 sqrt(4-x)=12
sqrt(4-x)=4
Now, to solve the above equation we can raise both sides of the given equation to the power of 2.
sqrt(4-x)=4
( sqrt(4-x) )^2=4^2
4-x=4^2
â–Ľ
Solve for x
4-x=16
- x=12
x=- 12
Raising both sides of the equation to the power of 2, we found that the solution of the given equation is x=- 12. Let's represent it on a number line!
f We are asked to solve the given inequality and graph the solution set on the number line.
|x+1| ≤ 4 To do this we will create a compound inequality by removing the absolute value. In this case, the solution set is any number less than or equal to 4 away from the midpoint in the positive direction and any number less than or equal to 4 away from the midpoint in the negative direction.

Absolute Value Inequality:& |x+1| ≤ 4 Compound Inequality:& - 4≤ x+1 ≤ 4 We can split this compound inequality into two cases, one where x+1 is greater than or equal to -4 and one where x+1 is less than or equal to 4. x+1 ≥ - 4 and x+1 ≤ 4 Let's isolate x in both of these cases before graphing the solution set.

Case 1

x+1 ≤ 4
x ≤ 3
This inequality tells us that all values less than or equal to 3 will satisfy the inequality.

Case 2

- 4 ≤ x+1
- 5 ≤ x
This inequality tells us that all values greater than or equal to - 5 will satisfy the inequality.

Solution Set

The solution to this type of compound inequality is the overlap of the solution sets. Let's recombine our cases back into one compound inequality. First Solution Set:& x ≤ 3 Second Solution Set:& - 5 ≤ x Intersecting Solution Set:& - 5 ≤ x ≤ 3

Graph

The graph of this inequality includes all values from - 5 to 3, inclusive. We show this by using closed circles on the endpoints.