How do you add #\frac { x } { x + 3} + \frac { 6x - 2} { x - 3}#?

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1 Answer
sente
Dec 29, 2016

#x/(x+3)+(6x-2)/(x-3)=(7x^2+13x-6)/(x^2-9)#

Explanation:

Step 1: Find the least common multiple of all denominators

#x+3# and #x-3# have no shared factors, so their least common multiple is their product: #(x+3)(x-3)#

Step 2: Multiply each term by an appropriate #1# to give them common denominators

#x/(x+3) = x/(x+3) * (x-3)/(x-3)= (x(x-3))/((x+3)(x-3))#

#(6x-2)/(x-3) = (6x-2)/(x-3) * (x+3)/(x+3) = ((6x-2)(x+3))/((x+3)(x-3))#

Step 3: Add the numerators of the new fractions

#x/(x+3)+(6x-2)/(x-3) = (x(x-3))/((x+3)(x-3))+((6x-2)(x+3))/((x+3)(x-3))#

#=(x(x-3)+(6x-2)(x+3))/((x+3)(x-3))#

Step 4: Simplify

#(x(x-3)+(6x-2)(x+3))/((x+3)(x-3)) = ((x^2-3x)+(6x^2+16x-6))/(x^2-9)#

#=(7x^2+13x-6)/(x^2-9)#

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