Question

(2x-4)/(4x^2 -25) - (1)/(4x-10) = (5)/ (6x+15)

Answer 1

HI :DD

Answer 2

Hi. Can you help?

Answer 3

@satellite73

Answer 4

hi

Answer 5

Can you please help??

Answer 6

\[\frac{2x-4}{4x^2 -25} - \frac{1}{4x-10} = \frac{5}{6x+15}\]i just needed to write it first
is that right?

Answer 7

Yesss

Answer 8

i was afraid so
ok lets factor the denominators

Answer 9

\[\frac{2x-4}{4x^2 -25} - \frac{1}{4x-10} = \frac{5}{6x+15}\]
\[\frac{2x-4}{(2x+5)(2x-5)} - \frac{1}{2(x-5)} = \frac{5}{3(2x+5)}\]

Answer 10

Wouldn't the second one be 2(2x-5)

Answer 11

yes, you are right

Answer 12

\[\frac{2x-4}{(2x+5)(2x-5)} - \frac{1}{2(2x-5)} = \frac{5}{3(2x+5)}\]

Answer 13

I did that. After that I got lost

Answer 14

least common multiple of the denominators is \(6(2x+5)(2x-5)\) so we can clear the fractions if we multiply both sides by it, cancelling as we go

Answer 15

\[6(2x+5)(2x-5)\left(\frac{2x-4}{(2x+5)(2x-5)} - \frac{1}{2(2x-5)} \right)\]\[=6(2x+5)(2x-5)\left( \frac{5}{3(2x+5)}\right)\]

Answer 16

just a lot of cancellation here
you get no fractions now
\[6(2x-4)-3(2x+5)=2\times 5(2x-5)\]

Answer 17

now it should be a lot easier
multiply out, combine like terms, etc
gives
\[12x-24-6x-15=20x-50\] and so on
you good from there?

Answer 18

Can you show me how you canceled them out???

Answer 19

\[6(2x+5)(2x-5)\left(\frac{2x-4}{(2x+5)(2x-5)} - \frac{1}{2(2x-5)} \right)\]
\[6(2x+5)(2x-5)\left(\frac{2x-4}{(2x+5)(2x-5)}\right) -6(2x+5)(2x-5) \left(\frac{1}{2(2x-5)}\right)\]
\[6\cancel{(2x+5)(2x-5)}\left(\frac{2x-4}{\cancel{(2x+5)(2x-5)}}\right)\]\[ -\cancel{6}^3(2x+5)\cancel{(2x-5)} \left(\frac{1}{\cancel{2(2x-5)}}\right)\]

\[=6(2x-4)-3(2x+5)\]

Answer 20

got rid of all the common factors top and bottom

Answer 21

kind of long to draw out, and that was just the left hand side

Answer 22

I get it now!!!!

Answer 23

great! get rid of common factors, end up with a nice linear equation to solve
but you still have to be careful

Answer 24

in any case if you finish solving the last equation i think you get \(\frac{11}{14}\)
let me know if it works out