Question 846547
{{{2((x-1)/(x+3))-7((x+3)/(x-1))=5}}}

The first thing we need to do is multiply 2 by (x - 1) and multiply 7 by (x + 3):


{{{((2x-2)/(x+3))-((7x+21)/(x-1))=5}}}

Next, we need to subtract the remaining fractions on the left side of the equation.  We do this by multiplying (2x - 2) by (x - 1) and multiplying (7x + 21) by (x + 3) and setting this result in the numerator and setting (x + 3)(x - 1) in the denominator:


{{{(((2x-2)*(x-1))/((x+3)*(x-1)))-(((7x+21)*(x+3))/((x+3)*(x-1)))=5}}} ----->


{{{(((2x^2-4x+2))/((x+3)*(x-1))-((7x^2+42x+63))/((x+3)*(x-1)))=5}}}


Next, after distributing the minus sign to (7x^2 + 42x + 63), combine all of the terms in the numerator and set that above the denominator:


{{{(2x^2-4x+2-7x^2-42x-63)/((x+3)(x-1))=5}}} ----->


{{{(-5x^2-46x-61)/((x+3)(x-1))=5}}}


Next, since this single fraction is set equal to a number, we will cross-multiply the numerator in the fraction, by 1, since 5 = 5/1, and multiply the denominator in the fraction by 5:


{{{-5x^2-46x-61=5(x+3)(x-1)}}} ----->


{{{-5x^2-46x-61=5(x^2+2x-3)}}} ----->


{{{-5x^2-46x-61=5x^2+10x-15}}}


Now, since this is a quadratic equation, we need to move all of the terms to one side of the equation and set the other side of the equation equal to 0:


{{{0=5x^2+5x^2+10x+46x-15+61}}}


Combining like terms gives us:


{{{0=10x^2+56x+46}}}


Now we have this in quadratic form.  Let's make this equation easier to factor by factoring out 2, giving us:


{{{0=2(5x^2+28x+23)}}}


Now, this is easily factorable:


{{{0=2(5x+23)(x+1)}}}


We can discard the 2 we factored out, and set each of our factors equal to zero to find our x's:


{{{5x+23=0}}} ----->


{{{5x=-23}}} ----->


{{{x=-23/5}}}


AND


{{{x+1=0}}} ----->


{{{x=-1}}}


We now have solved for x.


x = -1 and -23/5


You need to verify both of these values of x work by plugging each value of x into the original equation.  Both, in fact, do work, so


x = -1 and -23/5