Question 701553
First thing in problems like this is to get rid of those horrible fractions!
{{{(7/(2x+1))-(8x/(2x-1))=-4}}}
If we multiply each term through the equation by (2x+1) we get
{{{(2x+1)(7/(2x+1))-(2x+1)(8x/(2x-1))=(2x+1)(-4)}}} the 2x+1 in the denominator of the first term will cancel with the 2x+1 on the top line
giving {{{7-(2x+1)(8x/(2x-1))=(2x+1)(-4)}}}
that's got rid of the fraction from the first term so lets do a similar thing to the get rid of the second fraction.
This time we'll multiply each term by 2x-1 this will cancel with the denominator in the second fraction to give
{{{(2x-1)(7)-(2x+1)(8x)=(2x+1)(2x-1)(-4)}}}
Now we can multiply out all those brackets gives
{{{14x-7-16x^2-8x=-16x^2+4}}}
Simplifying gives
6x=3 therefore dividing by 6 gives
x={{{1/2}}}