Question

question 9 of 10, step 1 of 1
solve the following rational equation and simplify your answer.
\\(\frac{z^3 - 3z^2}{z^2 + 6z - 27} = \frac{-11z - 18}{z + 9}\\)
answer
\\(z = \\)

Explanation:

Step1: Factor all polynomials

Left numerator: $z^3 - 3z^2 = z^2(z - 3)$
Left denominator: $z^2 + 6z - 27 = (z + 9)(z - 3)$
Right numerator: $-11z - 18$ (cannot be factored)
Right denominator: $z + 9$ (cannot be factored)
Equation becomes: $\frac{z^2(z - 3)}{(z + 9)(z - 3)} = \frac{-11z - 18}{z + 9}$

Step2: Cancel common factors

Cancel $(z - 3)$ (where $z
eq 3$) and $(z + 9)$ (where $z
eq -9$):
$z^2 = -11z - 18$

Step3: Rearrange to standard quadratic

$z^2 + 11z + 18 = 0$

Step4: Factor quadratic

$(z + 2)(z + 9) = 0$

Step5: Solve for z and exclude extraneous roots

Solutions from factoring: $z = -2, z = -9$
Exclude $z = -9$ (makes original denominators 0)

Answer:

$z = -2$