Solve {c}{w^2(x+8)+z(x+8)=(quad)(x+8)}{text{Enterthecorrectanswer.}}

$$\left. \begin{array} { c } { w ^ { 2 } ( x + 8 ) + z ( x + 8 ) = ( \quad ) ( x + 8 ) } \\ { \text { Enter the correct answer. } } \end{array} \right.$$

Solve for x

$$\left\{\begin{matrix}\\x=-8\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&z=1-w^{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply w^{2} by x+8.

$$w^{2}x+8w^{2}+z\left(x+8\right)=x+8$$

Use the distributive property to multiply z by x+8.

$$w^{2}x+8w^{2}+zx+8z=x+8$$

Subtract x from both sides.

$$w^{2}x+8w^{2}+zx+8z-x=8$$

Subtract 8w^{2} from both sides.

$$w^{2}x+zx+8z-x=8-8w^{2}$$

Subtract 8z from both sides.

$$w^{2}x+zx-x=8-8w^{2}-8z$$

Combine all terms containing x.

$$\left(w^{2}+z-1\right)x=8-8w^{2}-8z$$

The equation is in standard form.

$$\left(z+w^{2}-1\right)x=8-8w^{2}-8z$$

Divide both sides by w^{2}+z-1.

$$\frac{\left(z+w^{2}-1\right)x}{z+w^{2}-1}=\frac{8-8w^{2}-8z}{z+w^{2}-1}$$

Dividing by w^{2}+z-1 undoes the multiplication by w^{2}+z-1.

$$x=\frac{8-8w^{2}-8z}{z+w^{2}-1}$$

Divide 8-8w^{2}-8z by w^{2}+z-1.

$$x=-8$$