Answer:
[tex]\frac{x-1}{x+3}[/tex]
Step-by-step explanation:
Let's factor the numerator and denominator first.
x^2+5x-6 is a quadratic in the form of x^2+bx+c.
If you have a quadratic in the form of x^2+bx+c, all you have to do to factor is think of two numbers that multiply to be c and add to be b.
In this case multiplies to be -6 and adds to be 5.
Those numbers are 6 and -1 since -1(6)=-6 and -1+6=5.
So the factored form of x^2+5x-6 is (x-1)(x+6).
x^2+9x+18 is a quadratic in the form of x^2+bx+c as well.
So we need to find two numbers that multiply to be 18 and add to be 9.
These numbers are 6 and 3 since 6(3)=18 and 6+3=9.
So the factored form of x^2+9x+18 is (x+3)(x+6).
So we have that:
[tex]\frac{x^2+5x+-6}{x^2+9x+18}=\frac{(x-1)(x+6)}{(x+3)(x+6)}[/tex]
We can simplify this as long as x is not -6 as
[tex]\frac{x-1}{x+3}[/tex]
I obtained the last line there by canceling out the common factor on top and bottom.
