How do you simplify #\frac { 4} { y - 5} - \frac { 6} { y + 5} + \frac { 2y } { y ^ { 2} - 25}#?

2 Answers
Tom M.
Nov 21, 2017

#50/(y^2-25)#

Explanation:

We'll begin by evaluating #4/(y-5)-6/(y+5)#

To subtract these, we need to put them over a common denominator. So we'll multiply top and bottom of the first fraction by #(y+5)# (also know as the conjugate of #(y+5)#), and top and bottom of the second by #(y-5)#

#(4)/((y-5))-(6)/((y+5))+(2y)/(y^2-25)#

#-=(4(y+5))/((y-5)(y+5))-(6(y-5))/((y+5)(y-5))+...#

Now we have a common denominator, we can subtract these fractions.

#-=(4(y+5)-6(y-5))/((y+5)(y-5))+...#
#-=(4y+20-6y+30)/(y^2-25)+...#
#-=(50-2y)/(y^2-25)+(2y)/(y^2-25)#
Since we have already have a common denominator, we can add these fractions.
#-=(50-2y+2y)/(y^2-25)#
#-=50/(y^2-25)#

smendyka
Nov 21, 2017

See a solution process below:

Explanation:

Use this rule of quadratics to find the factors of the fraction on the right:

#color(red)(x)^2 - color(blue)(y)^2 = (color(red)(x) + color(blue)(y))(color(red)(x) - color(blue)(y))#

#color(red)(y)^2 - color(blue)(25) => color(red)(y)^2 - color(blue)(5)^2 => (color(red)(y) + color(blue)(5))(color(red)(y) - color(blue)(5))#

We can now multiply the two fractions on the left by the appropriate form of #1# to put each fraction over a common denominator:

#((y + 5)/(y + 5) xx 4/(y - 5)) - ((y - 5)/(y - 5) xx 6/(y + 5)) + (2y)/((y + 5)(y -5))#

#(4(y + 5))/((y + 5)(y - 5)) - (6(y - 5))/((y + 5)(y - 5)) + (2y)/((y + 5)(y -5))#

#(4y + 20)/((y + 5)(y - 5)) - (6y - 30)/((y + 5)(y - 5)) + (2y)/((y + 5)(y -5))#

We can now add and subtract the numerators over the common denominator:

#((4y + 20) - (6y - 30) + 2y)/((y + 5)(y -5))#

#(4y + 20 - 6y + 30 + 2y)/((y + 5)(y -5))#

#(4y - 6y + 2y + 20 + 30)/((y + 5)(y -5))#

#((4 - 6 + 2)y + (20 + 30))/((y + 5)(y -5))#

#(0y + 50)/((y + 5)(y -5))#

#50/((y + 5)(y -5))#

Or

#50/(y^2 - 25)#

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