Question 95951
Given to solve:
.
{{{(7/(2x+1))-(8x/(2x-1))=-4}}}
.
If you don't like fractions, in this case you can get rid of the fractions. How? Begin by noting
that {{{(2x+1)(2x-1)}}} is a common denominator. Suppose that we multiply both sides of this
equation (all terms) by {{{(2x+1)(2x-1)}}}. In doing that multiplication you make the original
problem become:
.
{{{((2x+1)(2x-1)*7)/(2x+1) -((2x+1)(2x-1)*8x)/(2x-1) = -4*(2x+1)(2x-1)}}}
.
Next, where it applies you can cancel the denominator with the like term in the numerator as
follows:
.
{{{((cross(2x+1))*(2x-1)*7)/(cross(2x+1)) -((2x+1)(cross(2x-1))*8x)/(cross(2x-1)) = -4*(2x+1)(2x-1)}}}
.
After these cancellations you are left with:
.
{{{((2x -1)*7) - ((2x+1)*8x) = -4*(2x+1)(2x-1)}}}
.
Multiplying {{{(2x - 1)*7}}} results in {{{14x - 7}}}
.
Multiplying {{{(2x + 1)*8x}}} results in {{{16x^2 + 8x}}}
.
and finally, 
.
Multiplying {{{-4*(2x+1)(2x -1)}}} results in {{{-16x^2 + 4}}}
.
Substitute these results into the equation and you have:
.
{{{14x - 7 - (16x^2 + 8x) = -16x^2 + 4}}}
.
On the left side you can remove the parentheses if you change the signs of the terms inside
the parentheses to get:
.
{{{14x - 7 - 16x^2 - 8x = -16x^2 + 4}}}
.
Notice that you have {{{-16x^2}}} on both sides. Therefore, if you add {{{16x^2}}} to
both sides these two terms disappear and you are left with:
.
{{{14x - 7 - 8x = 4}}}
.
On the left side you can combine the {{{14x}}} and the {{{-8x}}} to get {{{6x}}} to reduce
the equation to:
.
{{{6x - 7 = 4}}}
.
Then get rid of the -7 on the left side by adding 7 to both sides to get:
.
{{{6x = 11}}}
.
Finally, solve for x by dividing both sides by 6 to get:
.
{{{x = 11/6}}}
.
This is the answer you are looking for ... {{{x = 11/6}}}
.
You can check this answer by substituting {{{11/6}}} for x in the original problem and
ensuring that the equation will balance. 
.
Hope this helps you to understand the problem and how you can get rid of the denominators in
the equation so that you are not working with fractions.
.