Core Connections Algebra 1, 2013
CC
Core Connections Algebra 1, 2013 View details
3. Section 10.3
Continue to next subchapter

Exercise 141 Page 515

Practice makes perfect
a This equation would be much easier to solve if it had smaller coefficients. We can start solving by changing this equation to a simpler equivalent equation by dividing both sides of the equation by 300.
300x-1500=2400
x-5=8
To solve the above equation, we should gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality.
x-5=8
x=13
The solution to the equation is x=13. We can check our solution by substituting it into the original equation.
300x-1500=2400
300( 13)-1500 ? = 2400
â–Ľ
Simplify
3900-1500 ? = 2400
2400=2400 âś“
Since the left-hand side is equal to the right-hand side, our solution is correct.
b This equation would be much easier to solve if it had no fractions. We can start solving by changing this equation to a simpler equivalent equation by eliminating fractions. To do this, we will multiply both sides of the equation by 6, which is the lowest common denominator.
3/2x=5/6x+2
18/2x=30/6x+12
9x=5x+12
To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality.
9x=5x+12
4x=12
x=3
The solution to the equation is x=3. We can check our solution by substituting it into the original equation.
3/2x=5/6x+2
3/2( 3)? = 5/6( 3)+2
â–Ľ
Simplify
9/2? = 15/6+2
9/2? = 5/2+2
9/2? = 5/2+4/2
9/2=9/2 âś“
Since the left-hand side is equal to the right-hand side, our solution is correct.
c To solve the given inequality we will use square roots. To do this, we need to gather all variable terms on one side of the inequality and all constant terms on the other side. Let's do this!
x^2-25 ≤ 0 ⇔ x^2 ≤ 25 Now we can apply the square root to both sides of the inequality. Remember that we do not know the sign of the variable x, so we need to use the absolute value.
x^2 ≤ 25
sqrt(x^2) ≤ sqrt(25)
|x| ≤ 5
To solve this inequality 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 5 away from the midpoint in the positive direction and any number less than 5 away from the midpoint in the negative direction. Absolute Value Inequality:& |x| ≤ 5 Compound Inequality:& - 5≤ x ≤ 5
d We are asked to solve the given inequality.
|3x-2|> 4

To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set contains the numbers that make the distance between 3x and 2 greater than 4 in the positive direction or in the negative direction. 3x-2 > 4 or 3x-2 <- 4 To find the solution, let's isolate x in both of these cases.

Case 1

3x-2 > 4
3x > 6
x>2
This inequality tells us that all values greater than 2 will satisfy the inequality.

Case 2

3x-2 <- 4
3x < - 2
x< - 2/3
x< - 2/3
This inequality tells us that all values less than - 23 will satisfy the inequality.

Solution Set

The solution to this type of compound inequality is the combination of the solution sets. First Solution Set:& x>2 Second Solution Set:& x < - 23 Combined Solution Set:& x < - 23 or x>2