Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
1. Section 4.1
Continue to next subchapter

Exercise 53 Page 184

Practice makes perfect
a The equation can be solved if we distribute 3 to each term in the parentheses and then use inverse operations to isolate x.
5-3(1/2x+2)=-7
5-3(x/2+2)=-7
5-3(x/2)-6=-7
5-3x/2-6=-7
5-6=-7+3x/2
6=3x/2
12=3x
3x=12
x=4
b The equation is a radical equation. It can be solved by inverse operations. Notice though that to solve for x, we have to square both sides which can introduce extraneous solutions.
5(sqrt(x-2)+1)=15
sqrt(x-2)+1=3
sqrt(x-2)=2
x-2=4
x=6
The solution to the equation is x=6. However, we need to check if it's an extraneous solution. To do this, we substitute x with 6 in the equation and see if the equality holds.
5(sqrt(x-2)+1)? =15
5(sqrt(6-2)+1)? =15
â–Ľ
Evaluate left-hand side
5(sqrt(4)+1)? =15
5(2+1)? =15
5(3)? =15
15=15 âś“
Since the equality is true, x=6 is a solution to the equation.
c The first step is to remove the parentheses. Since there is a minus sign in front of it, we have to change the signs inside the parentheses when removing it. We can then solve the equation with inverse operations.
12-(2x/3+x)=2
12-2x/3-x=2
12-2x/3-3x/3=2
12+- 2x/3-3x/3=2
12+-5x/3=2
-5x/3=-10
-5x=-30
x=6
d The equation is cubic, and it can be solved by taking the cube root of both sides. However, we must first isolate the cubed expression.
-3(2x+1)^3=-192
3(2x+1)^3=192
(2x+1)^3=64
2x+1=4
2x=3
x=3/2