this is part of a test, so if you could tell me "how" to do this that would be great!

(x/5)-(x+3/3)=(1/15)

like I said in the title, this is a fractional equation, I keep getting x=7 but it will not come out right when I check it.

any thoughts???

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- March 30th 2012, 02:14 PMDrummermanfraction equation
this is part of a test, so if you could tell me "how" to do this that would be great!

(x/5)-(x+3/3)=(1/15)

like I said in the title, this is a fractional equation, I keep getting x=7 but it will not come out right when I check it.

any thoughts??? - March 30th 2012, 02:33 PMemakarovRe: fraction equation
First, I assume the equation is x/5 - (x + 3)/3 = 1/15 (note the parentheses around x + 3; otherwise x + 3/3 = x + (3/3) = x + 1).

Multiply both sides by 15. Post the result and if you still have a difficulty, describe it. - March 30th 2012, 10:26 PMDrummermanRe: fraction equation
yes emakarov, sorry about the error. you could say it as "x plus three over three" and I went that route, I will post all of my work and see if you know where my flaw is.

x/5 - (x+3)/3 = 1/15

15(x/5) - 15([x+3]/3) = 15(1/15)

15x/5 - (15x+45)/3 = 1

3x - (5x+15) = 1

(-2x) + 15 = 1

15 = 2x + 1

14 = 2x

7 = x

That looks more confusing on paper... lol but that is what I keep coming to (that 7 = x) and when I use 7 = x and check it I end up with 29/15 = 1/15 (which is quite obviously wrong... LOL) - March 30th 2012, 11:15 PMbiffboyRe: fraction equation
Your 5th line should be -2x-15=1

- March 31st 2012, 08:58 AMDrummermanRe: fraction equation
OH!!! thank you! I guess I pulled down the + instead... I got it now!!!