Question

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

Answer 1

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}`