- Thread starter fiskonian
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Show your work - we can find the problem.fiskonian said:I need help with this problem.

1/9(x+18)+1/3(2x+3)=x+3

I know the answer is 0 because i looked in the back of the book to check my work.

However i cannot seem to work it out to get 0. please help me.

By the way, the answer in the book is correct.

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Working backwards from the answer, I think you mean:fiskonian said:I need help with this problem.

1/9(x+18)+1/3(2x+3)=x+3

. . . . .(1/9)(x + 18) + (1/3)(2x + 3) = x + 3

After you multiplied through by the common denominator, and simplified the parenthetical stuff, what did you do next? Please reply showing all of your steps.

Thank you!

Eliz.

P.S. Welcome to FreeMathHelp!

1/9(x+18)+1/3(2x+3)=x+3

9(1/9[x+18]+1/3[2x+3]=x+3)

x+18+6x+9=x+27

7x+27=x+27 (once i got to here i knew the answer... finding out how to get to there was what i had problems with.)

6x=0

x=0

im not sure if this is followable, its hard to show my work on this. PS- ill probably be on here alot these next few years

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Hint: 9(x) does not equal x. :wink:fiskonian said:1/9(x+18)+1/3(2x+3)=x+3

9(1/9[x+18]+1/3[2x+3]=x+3)

x+18+6x+9=x+27

Eliz.

The common denominator for the fractions is 9. Multiply both sides of the equation by 9:fiskonian said:

1/9(x+18)+1/3(2x+3)=x+3

9(1/9[x+18]+1/3[2x+3]=x+3)

x+18+6x+9=x+27

7x+27=x+27 (once i got to here i knew the answer... finding out how to get to there was what i had problems with.)

6x=0

x=0

im not sure if this is followable, its hard to show my work on this. PS- ill probably be on here alot these next few years

9*(1/9)(x + 18) + 9*(1/3)(2x + 3) = 9*(x + 3) <======here's where your mistake is!!

1(x + 18*) + 3(2x + 3) = 9(x + 3)

x + 18 + 6x + 9 = 9x + 27

Ok...now see what this gives you!

First of all, that should be shown this way: (x + 18) / 9 + (2x + 3) / 3 = x + 3

Then the multiplication by 9 affects EACH of the terms:

9 * (x + 18) / 9 = x + 18

9 * (2x + 3) / 3 = 3(2x + 3) = 6x + 9

9 * (x + 3) = 9x + 27

Kapish :?: