How do you combine #4/(x^2+4x-5) - 3/(x^2-1)#?

2 Answers
May 19, 2015

We can factor both denominators in order to find the lowest common denominator among them.

Using Bhaskara for the first equation #x^2+4x-5#

#(-4+-sqrt(16-4(1)(-5)))/2#
#(-4+-6)/2#

#x_1=-5#, which is the same as the factor #x+5=0#
#x_2=1#, which is the same as the factor #x-1=0#

For the second equation, we can find the roots simply by equaling it to zero.

#x^2-1=0#
#x^2=1#
#x=sqrt(1)#

#x_1=1#, which is the same as the factor #x-1=0#
#x_1=-1#, which is the same as the factor #x+1=0#

Now, we can rewrite your subtraction as:

#4/((x+5)(x-1))-3/((x-1)(x+1))#

The lowest common denominator must comprise all terms in the denominators, which lead us to #(x+5)(x-1)(x+1)#, where all the elements of both denominators are included and, thus, we can proceed to subtract.

Now we know our denominator is #(x+5)(x-1)(x+1)#, let's complete the subtraction:

#(4(x+1)-3(x+5))/((x+5)(x-1)(x+1))#

If you want to redistribute, it'll end up like this:

#(x-11)/(x^3+4x^2-5)#

May 19, 2015

Factor #y = (x^2 + 4x - 5) = (x -1)(x + 5) #

#4/[(x - 1)(x + 5)] - 3/[(x - 1)(x + 1)]# =

Numerator: 4x + 4 - 3x - 15 = x - 11

# y = (x - 11)/[(x - 1)(x + 5)(x + 1)]#