Question

How do you solve 2/(5x-5) + (x-2)/15 = 4/(5x-5)

Answer 1

\[\frac{2}{5x-5}+\frac{2-x}{15}=\frac{4}{5x-5}\]What if you moved the fraction \(\dfrac{2}{5x-5}\) to the right hand side, and then cross multiplied?

Answer 2

are you simplifying? if so, here's your answer:

(x^2-5x+6)(x^2-5x+5)=
x^4-5x^3+5x^2-5x^3+25x^2-25x+6x^2-30x+…
x^4-10x^3+36x^2-55x+30

Answer 3

@jasmyn_necoleee nope, nothing to do with the problem

Answer 4

The easiest way is multiply the two sides with 5x-5 so you get off x in the doniminator

Answer 5

im sorry i did the wrong problem i meant to post this somewhere else .. this is way of what you are doing .

Answer 6

After moving the fraction (which combines with the fraction already there), you have \[\frac{2-x}{15}=\frac{2}{5x-5}\]Cross-multiply
\[(2-x)(5x-5)=30\]Expand and solve with the quadratic formula.

If you multiply by (5x-5), you get
\[2+\frac{(5x-5)(x-2)}{15} = 4\]Move the 2, multiply by 15 to clear the fraction, and you've done the same amount of work.

Answer 7

\[2/(5x-5)+(2-x)/15=4/(5x-5)\]Subtract 2/(5x-5) from both sides.\[(2-x)/15=(4-2)/(5x-5)\]\[(2-x)/15=2/(5x-5)\]Multiply both sides by (5x-5)/2\[(2-x)(5x-5)/2*15=1\]\[(10x-10-5x^2+5x)/30=1\]\[-5x^2+15x-10=30\]\[5x^2-15x+10=-30\]\[x^2-3x+2=-6\]\[x^2-3x+8=0\]
There are no real solutions.

Answer 8

Thanks

Answer 9

Uh, other than x = -1 and x = 4? :-)

We all must be making a mistake somewhere, because x = 4 certainly solves the original problem.

Answer 10

Strange, I certainly don't see where the error is in the approach that @mitodoteira and I took, but it doesn't produce the correct answer...the solutions aren't restricted values of the fractions. a puzzlement. @satellite73