168k views
3 votes
Question 3 of 8 Solve the equation below. 6/x^2+2x-15+7/x+5=2/x-3

1 Answer

5 votes

I assume the equation is


\frac6{x^2+2x-15}+\frac7{x+5}=\frac2{x-3}

Notice that


x^2+2x-15=(x+5)(x-3)

so to get each fraction to have a common denominator, we need to rewrite


\frac7{x+5}=(7(x-3))/((x+5)(x-3))=(7x-21)/(x^2+2x-15)

and


\frac2{x-3}=(2(x+5))/((x+5)(x-3))=(2x+10)/(x^2+2x-15)

So we have


\frac6{x^2+2x-15}+(7x-21)/(x^2+2x-15)=(2x+10)/(x^2+2x-15)

Combine the fractions and put them on one side:


(6+(7x-21)-(2x+10))/(x^2+2x-15)=0

If x ≠ -5 and x ≠ 3, we can ignore the denominator, leaving us with


6+(7x-21)-(2x+10)=0


(6-21-10)+(7x-2x)=0


-25+5x=0


5x=25


x=\frac{25}5=\boxed{5}

User David Andrei Ned
by
5.1k points