How do you simplify #\frac { x + 5y } { x + 3y } - \frac { 2x + y } { 3y + x } + \frac { 4x + 5y } { x + 3y }#?

1 Answer

#3#

Explanation:

Since all the denominators are equal, we can just combine the numerators together from
#(x+5y)/(x+3y)-(2x+y)/(3y+x)+(4x+5y)/(x+3y)# to

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

Let's work on the numerator first. First, we remove the brackets

#x+5y -2x-y +4x+5y#

Then group like terms.

#x-2x+4x+5y -y +5y#

#x+4x-2x =3x" "# and #" "5y-y+5y=9y#

The expression will now be #3x+9y#.

Put them all together as a fraction and you get #(3x+9y)/(x+3y)#.

Next, we will factor out the common term, which is #3#, to get

#(3(x+3y))/(x+3y)#.

We will cancel out #x+3y#.

#(3(cancel(x+3y)))/(cancel(x+3y))=3#.

The answer is #3#.