Question 547547
{{{ 3/2x}}}{{{ - x/4x^2}}}{{{-1 = 7/4x+2}}}? That would be way to simple.
You probably meant 3/2x - x/(4x^2-1) = 7/(4x+2), which I can write as
{{{3/2x - x/(4x^2-1) = 7/(4x+2)}}}.
The first thing to do is looking for factors in the denominators. I can factor them, to get:
{{{3/2x - x/((2x-1)(2x+1)) = 7/(2(2x+1))}}}
A good common denominator would be {{{2x(2x-1)(2x+1)}}}, but instead of writing denominators over and over I will multiply both sides of the equation by that expression to eliminate denominators.
{{{3*cross(2x)(2x-1)(2x+1)/cross(2x) - x*2x*cross((2x-1)(2x+1))/cross((2x-1)(2x+1)) = 7*cross(2)*x*(2x-1)*cross(2x+1)/(cross(2)*cross(2x+1))}}}, which simplifies to
{{{3*(2x-1)(2x+1) - x*2x = 7*x*(2x-1)}}}
Multiplying as indicated, we get
{{{12x^2-3 - 2x^2 = 14x^2-7x}}} --> {{{10x^2-1 = 14x^2-7x}}}
Adding {{{3-10x^2}}} to both sides, we get {{{0=4x^2-7x+3}}}
Whichever way you solve that quadratic equation (factoring, completing the square, or applying the quadratic formula), you get the two solutions
{{{x=1}}} and {{{x=3/4}}}
It's a good idea to check, because, even if we make no mistakes, on eliminating denominators we can introduce extraneous solutions, that were not solutions of the original equation because they made one of the denominators zero. Our solutions do not make any of the denominators zero, but I often make mistakes, so I'll verify the solutions.
For {{{x=1}}}, {{{3/2x - x/(4x^2-1) =3/(2*1) - 1/(4*1^2-1)=3/2-1/3=7/6}}} and {{{7/(4x+2)=7/(4*1+2)=7/6}}}
For {{{x=3/4}}}, {{{3/2x - x/(4x^2-1) =3/(2*(3/4)) - (3/4)/(4*(3/4)^2-1)=3/(3/2)-(3/4)/(9/4-1)=2-(3/4)/(5/4)=2-3/5=7/5}}} and {{{7/(4x+2)=7/(4*(3/4)+2)=7/(3+2)=7/5}}}
Both solutions work.