Solve (9x+3y)+2z=9x+left(3y+2zright)qu

$$(9x+3y)+2z = 9x+ \left( 3y+2z \right) qu$$

Solve for q

$$\left\{\begin{matrix}q=\frac{1}{u}\text{, }&u\neq 0\\q\in \mathrm{R}\text{, }&y=-\frac{2z}{3}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 3y+2z by q.

$$9x+3y+2z=9x+\left(3yq+2zq\right)u$$

Use the distributive property to multiply 3yq+2zq by u.

$$9x+3y+2z=9x+3yqu+2zqu$$

Swap sides so that all variable terms are on the left hand side.

$$9x+3yqu+2zqu=9x+3y+2z$$

Subtract 9x from both sides.

$$3yqu+2zqu=9x+3y+2z-9x$$

Combine 9x and -9x to get 0.

$$3yqu+2zqu=3y+2z$$

Combine all terms containing q.

$$\left(3yu+2zu\right)q=3y+2z$$

The equation is in standard form.

$$\left(3uy+2uz\right)q=3y+2z$$

Divide both sides by 3yu+2zu.

$$\frac{\left(3uy+2uz\right)q}{3uy+2uz}=\frac{3y+2z}{3uy+2uz}$$

Dividing by 3yu+2zu undoes the multiplication by 3yu+2zu.

$$q=\frac{3y+2z}{3uy+2uz}$$

Divide 3y+2z by 3yu+2zu.

$$q=\frac{1}{u}$$

Solve for u

$$\left\{\begin{matrix}u=\frac{1}{q}\text{, }&q\neq 0\\u\in \mathrm{R}\text{, }&y=-\frac{2z}{3}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 3y+2z by q.

$$9x+3y+2z=9x+\left(3yq+2zq\right)u$$

Use the distributive property to multiply 3yq+2zq by u.

$$9x+3y+2z=9x+3yqu+2zqu$$

Swap sides so that all variable terms are on the left hand side.

$$9x+3yqu+2zqu=9x+3y+2z$$

Subtract 9x from both sides.

$$3yqu+2zqu=9x+3y+2z-9x$$

Combine 9x and -9x to get 0.

$$3yqu+2zqu=3y+2z$$

Combine all terms containing u.

$$\left(3yq+2zq\right)u=3y+2z$$

The equation is in standard form.

$$\left(3qy+2qz\right)u=3y+2z$$

Divide both sides by 3yq+2zq.

$$\frac{\left(3qy+2qz\right)u}{3qy+2qz}=\frac{3y+2z}{3qy+2qz}$$

Dividing by 3yq+2zq undoes the multiplication by 3yq+2zq.

$$u=\frac{3y+2z}{3qy+2qz}$$

Divide 3y+2z by 3yq+2zq.

$$u=\frac{1}{q}$$