SOLUTION: 8-2(x-3)<4(x-10)+8

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Question 218773: 8-2(x-3)<4(x-10)+8
Found 2 solutions by drj, MathTherapy:
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
8-2%28x-3%29%3C4%28x-10%29%2B8

Step 1. Multiply to get rid of parenthesis

8-2x%2B6%3C4x-40%2B8

14-x%3C4x-32

Step 2. Add x+32 to both sides of inequality

14-x%2Bx%2B32%3C4x-32%2Bx%2B32

52%3C4x

Step 3. Divide 4 to both sides of the inequality

52%2F4%3C4x%2F4

13%3Cx

Step 4. ANSWER 13%3Cx


I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J
http://www.FreedomUniversity.TV

Answer by MathTherapy(639) About Me  (Show Source):
You can put this solution on YOUR website!
8 - 2(x - 3) < 4(x - 10) + 8

8 - 2x + 6 < 4x - 40 + 8 ----- Distribute to get rid of the parentheses

- 2x + 14 < 4x - 32

- 2x - 4x < - 32 - 14

- 6x < - 46

x > %28-46%29%2F-6, or x > 23%2F3, or, x > highlight_green%287.667%29. You should have realized that the < changes to > because of the divsion by a negative number

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Let's check
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Since x > 7.667, let's make x = 8

We then have:

8 - 2(x - 3) < 4(x - 10) + 8

8 - 2(8 - 3) < 4(8 - 10) + 8

8 - 2(5) < 4(- 2) + 8

8 - 10 < - 8 + 8

- 2 < 0 (TRUE)