Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
1. Section 11.1
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Exercise 5 Page 526

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 100.
900x-200=500x+600
9x-2=5x+6
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.
9x-2=5x+6
9x=5x+8
4x=8
x=2
The solution to the equation is x=2.
b We will use the Quadratic Formula to solve the given quadratic equation.
ak^2+ bk+ c=0 ⇕ k=- b± sqrt(b^2-4 a c)/2 aWe first need to identify the values of a, b, and c. 3k^2-15k+14=0 ⇕ 3k^2+( - 15)k+ 14=0 We see that a= 3, b= - 15, and c= 14. Let's substitute these values into the Quadratic Formula.
k=- b±sqrt(b^2-4ac)/2a
k=- ( - 15)±sqrt(( - 15)^2-4( 3)( 14))/2( 3)
Solve for k and Simplify
k=15±sqrt((- 15)^2-4(3)(14))/2(3)
k=15±sqrt(225-4(3)(14))/2(3)
k=15±sqrt(225-168)/6
k=15±sqrt(57)/6
k ≈ 15± 7.55/6
The solutions for this equation are k ≈ 15± 7.556. Let's separate them into the positive and negative cases.
k ≈ 15± 7.55/6
k_1 ≈ 15+ 7.55/6 k_2 ≈ 15- 7.55/6
k_1 ≈ 22.55/6 k_2 ≈ 6.69/6
k_1 ≈ 3.76 k_2 ≈ 1.24

Using the Quadratic Formula, we found that the solutions of the given equation are k_1 ≈ 3.76 and k_2 ≈ 1.24.

c We are asked to solve the given inequality.
|x-4| < 6 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 6 away from the midpoint in the positive direction and any number less than 6 away from the midpoint in the negative direction.

Absolute Value Inequality:& |x-4| < 6 Compound Inequality:& - 6< x-4 < 6 We can split this compound inequality into two cases — one where x-4 is greater than -6 and one where x-4 is less than 6. x-4>- 6 and x-4 < 6 Let's isolate x in both of these cases before graphing the solution set.

Case 1

x-4<6
x<10
This inequality tells us that all values less than 10 will satisfy the inequality.

Case 2

- 6< x-4
- 2
This inequality tells us that all values greater than - 2 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 < 10 Second Solution Set:& - 2 < x Intersecting Solution Set:& - 2 < x < 10

d 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.
7/3+x/2=6x-1/6
42/3+6x/2=6x-1
14+3x=6x-1
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.
14+3x=6x-1
15+3x=6x
15=3x
3x=15
x=5
The solution to the equation is x=5.