Question 232845
I am assuming you want this solved for 0
{{{2x/(x^2-9) + (1-2x)/(x^2-4x+3)=0}}} So first we need a common denominator
We will start by factoring the current denominators.
{{{2x/(x+3)(x-3)+(1-2x)/(x-3)(x-1)=0}}}
Now for the common denominator.
{{{2x(x-1)/(x+3)(x-3)(x-1)+(1-2x)(x+3)/(x+3)(x-3)(x-1)=0}}} Now your denominator cannot equal to zero since you are not allowed to divide by 0. so x cannot equal to 1, 3, or -3.
Since the bottom cannot equal 0 the only time the fraction will equal 0 is when the numerator equals 0.
Now we multiply out the numerator.
{{{(2x^2-2x+x+3-2x^2-6x)/(x+3)(x-3)(x-1)=0}}}
Next we combine like terms.
{{{(-7x+3)/(x+3)(x-3)(x-1)=0}}}
Now we set the numerator equal to 0 and solve.
{{{-7x+3=0}}} Add 7x to both sides.
{{{7x=3}}} Divide both sides by 7.
{{{x=3/7}}}