How do you combine #[(2x + y)/(x - y)] - [(3x - y)/(x + y)] - [(5x + 2y)/(y^2 - x^2)]#?

Question

Genesis Allen · Accepted Answer

Given: #[color(red)((2x+y)/(x-y))] - [color(blue)((3x-y)/(x+y))] -[color(green)(5(x+2y)/(y^2-x^2))]# Step 1: determine an appropriate common denominator
Since
#(x^2-y^2)#
#= (x+y) * color(red)((x-y))#
#=(x-y) * color(blue)((x+y)) #
#=(-1) * color(green)((y^2-x^2))#
#x^2-y^2# would seem to be the obvious choice for a common denominator Step 2:  Evaluate the numerators by multiplying each by the factor needed to obtain the common denominator
#([color(red)((2x+y)) * (x+y)] - [color(blue)((3x-y)) * (x-y)] -[color(green)((5x+2y)) * (-1)])/((x^2-y^2)# It is probably easiest to evaluate each numerator term separately then recombine
#+color(red)((2x+y)) * (x+y) = +(2x^2+3xy+y^2)#
# -color(blue)((3x-y)) * (x-y) = -(3x^2-4xy+y^2)#
#-color(green)((5x+2y)) * (-1) = +(5x+2y)# Giving a numerator sum of
#-x^2+7xy+5x+2y# Step 3: Recombine as a final solution
#(-x^2+7xy+5x+2y)/(x^2-y^2)#

Abigail Allen · Answer

We can find the lowest common denominator for your fractions. In order to do that, we'd better factor the third fraction's denominator, as it's a squared product and the others have degree one (#x^1#).

By factoring laws, we know that if #a^2-b^2# is the result, then it comes from #(a-b)(a+b)#. Thus,

#y^2-x^2=(y-x)(y+x)#

We can see that this fits as l.c.d for the second fraction's denominator (due to the term #(x+y)#) but it doesn't really fit the first one's. However, we can see that the denominator of the first one is the very opposite of #(y-x)#, thus #(y-x)=(-1)(x-y)#

So, our l.c.d. will be #(-1)(x-y)(x+y)#

Now, rewriting the third fraction and then proceeding to the calculation:

#(2x+y)/(x-y)-(3x-y)/(x+y)-(5x+2y)/((-1)(x-y)(x+y)#

#((-1)(x+y)(2x+y)-(-1)(x-y)(3x-y)-(5x+2y))/((-1)(x-y)(x+y))#

Aggregating and simplifying signals and factors:

#(-(x+y)(2x+y)+(x-y)(3x-y)-5x-2y)/(y^2-x^2)#

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

#(5x^2-5x-xy+y^2-2y)/(y^2-x^2)#

#color(green)((5x(x-1)+y(y-x-2))/(y^2-x^2))#

Genesis Allen · Answer

undefined To combine the given expression, we need to find a common denominator for all the fractions involved. The common denominator in this case is (x - y)(x + y)(y^2 - x^2).

After obtaining the common denominator, we can simplify the expression by multiplying each term by the appropriate factors to eliminate the denominators.

The simplified expression is:

[(2x + y)(x + y)(y^2 - x^2) - (3x - y)(x - y)(y^2 - x^2) - (5x + 2y)(x - y)(x + y)] / [(x - y)(x + y)(y^2 - x^2)]