SOLUTION: Anne solved 5(2x)- 3 = 20x + 15 for x by first distributing 5 on the left side of the equation. She got the answer x = -3. However, when she substituted -3 into the original situat

Algebra ->  Equations -> SOLUTION: Anne solved 5(2x)- 3 = 20x + 15 for x by first distributing 5 on the left side of the equation. She got the answer x = -3. However, when she substituted -3 into the original situat      Log On


   



Question 943678: Anne solved 5(2x)- 3 = 20x + 15 for x by first distributing 5 on the left side of the equation. She got the answer x = -3. However, when she substituted -3 into the original situation for x, she saw that her answer was wrong. What did Anne do wrong, and what is the correct way?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The mistake was to distribute the 5 factor. This factor is only on ONE term and the left member should simple be 10x-3.

5%282x%29-3=20x%2B15
10x-3=-20x%2B15
10x%2B20x-3=15
30x-3=15
30x=18
x=18%2F30
highlight%28x=3%2F5%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Anne solved 5(2x)- 3 = 20x + 15 for x by first distributing 5 on the left side of the equation. She got the answer x = -3. However, when she substituted -3 into the original situation for x, she saw that her answer was wrong. What did Anne do wrong, and what is the correct way?
She seemed to have distributed the 5 on the entire left side of the equation instead of only distributing
the 5 to the 2x. This is what she might've done:
5(2x - 3) = 20x + 15
10x - 15 = 20x + 15
10x - 20x = 15 + 15
- 10x = 30
x+=+30%2F%28-+10%29, or x+=+-+3
Correct solution, since she was wrong, as well as the person who responded:
5(2x) - 3 = 20x + 15
10x - 3 = 20x + 15
10x - 20x = 15 + 3
- 10x = 18
x+=+18%2F%28-+10%29, or x+=+9%2F%28-+5%29, or highlight_green%28x+=+-+1.8%29
You can do the check!!
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