(x-3/5)-1= (x-5)/4

Printable View

- Sep 8th 2009, 07:13 PMjoshua17hernandezI need help solving this equation
(x-3/5)-1= (x-5)/4

- Sep 8th 2009, 07:51 PMRapha
Hello joshua17hernandez!

You use brackets, I hope you know, what they mean. So you want to solve

$\displaystyle (x - \frac{3}{5}) - 1 = \frac{x-5}{4}$, right? Then

$\displaystyle x - \frac{3}{5} - \frac{5}{5} = \frac{x}{4} - \frac{5}{4}$

$\displaystyle x - \frac{8}{5} = \frac{x}{4} - \frac{5}{4}$

$\displaystyle - \frac{8}{5} + \frac{5}{4} = \frac{x}{4}-x$

$\displaystyle - \frac{32}{20} + \frac{25}{20} = \frac{x}{4}-\frac{4x}{4}$

$\displaystyle -\frac{7}{20} = \frac{-3x}{4}$

$\displaystyle -\frac{7*4}{20} = -3x$

$\displaystyle -\frac{7}{5} = -3x$

$\displaystyle -(-\frac{7}{5*3}) = x$

$\displaystyle +\frac{7}{15} = x$

Do you understand?

Rapha - Sep 8th 2009, 08:35 PMjoshua17hernandez
Oh yes, I did use the bracket...but i used them the wrong way sorry i meant to write it like this...

__x-3__-1=__x-5__

5 4

wait the 4 is suppose to be under the x-5

this is my real problem i kept getting x=9

but my book says the answer is x=-7 and i don't seem to understand how they got that answer.

Can you help once again? - Sep 8th 2009, 08:46 PMpacman
you mean (x-3)/5 - 1 = (x-5)/4? ok

multiply both sides WITH 20, the product of 5 and 4.

20(x-3)/5 - 20 (1) = 20(x-5)/4, cancel common factors

4(x-3) - 20 = 5(x-5);

4x - 12 - 20 = 5x - 25

4x - 5x = -25 + 32

-x = 7

x = -7.

Maybe this IS THE ANSWER that you want - Sep 8th 2009, 08:50 PMjoshua17hernandez
Oh yeah i see what i did, i had multiplied 4 by x, -3 and -1. Forgetting that i was suppose to multiply the -1 by 20.

Thanks Alot