Solve 9/2y=40/3z+74/3

$$\frac{ 9 }{ 2 } y = \frac{ 40 }{ 3 } z+ \frac{ 74 }{ 3 }$$

Solve for y

$$y=\frac{80z+148}{27}$$

Steps for Solving Linear Equation

The equation is in standard form.

$$\frac{9}{2}y=\frac{40z+74}{3}$$

Divide both sides of the equation by \frac{9}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{9}{2}y}{\frac{9}{2}}=\frac{40z+74}{3\times \frac{9}{2}}$$

Dividing by \frac{9}{2} undoes the multiplication by \frac{9}{2}.

$$y=\frac{40z+74}{3\times \frac{9}{2}}$$

Divide \frac{40z+74}{3} by \frac{9}{2} by multiplying \frac{40z+74}{3} by the reciprocal of \frac{9}{2}.

$$y=\frac{80z+148}{27}$$

Solve for z

$$z=\frac{27y}{80}-\frac{37}{20}$$

Steps for Solving Linear Equation

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

$$\frac{40}{3}z+\frac{74}{3}=\frac{9}{2}y$$

Subtract \frac{74}{3} from both sides.

$$\frac{40}{3}z=\frac{9}{2}y-\frac{74}{3}$$

The equation is in standard form.

$$\frac{40}{3}z=\frac{9y}{2}-\frac{74}{3}$$

Divide both sides of the equation by \frac{40}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{40}{3}z}{\frac{40}{3}}=\frac{\frac{9y}{2}-\frac{74}{3}}{\frac{40}{3}}$$

Dividing by \frac{40}{3} undoes the multiplication by \frac{40}{3}.

$$z=\frac{\frac{9y}{2}-\frac{74}{3}}{\frac{40}{3}}$$

Divide \frac{9y}{2}-\frac{74}{3} by \frac{40}{3} by multiplying \frac{9y}{2}-\frac{74}{3} by the reciprocal of \frac{40}{3}.

$$z=\frac{27y}{80}-\frac{37}{20}$$