How do you solve #\frac { 4} { n - 9} + \frac { 1} { n + 2} = \frac { 3} { n ^ { 2} - 7n - 18}#?

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Answer 1 · 2017-09-20T20:46:06

#n = 4/5#

Well, to add together the terms on the left, you'd have to have a common denominator.

Multiply the 1st term by #(n + 2)/(n+2)#, and the second by #(n - 9)/(n-9)#

Giving:

#(4(n+ 2))/((n-9)(n+2)) + (n-9)/((n-9)(n+2))#

#= (4n + 8 + n - 9)/(n^2-7n - 18) = (5n-1)/(n^2-7n-18)#

...and you are told that that equals:

#3/(n^2-7n-18)#

...and the denominators are equal. Isn't that convenient. So you can disregard them and use only the numerators to solve for n.

#5n-1 = 3#

#5n = 4#

#n = 4/5#

GOOD LUCK!