Solve {l}{40x=3y}{z=x+y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 40 x = 3 y }\\ { z = x + y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{3a}{43}$$

$$y=\frac{40a}{43}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x+y 40x=3y a=z$$

Substitute x+y for z in the equation a=z.

$$a=x+y$$

Solve the second equation for y and the third equation for x.

$$y=\frac{40}{3}x x=a-y$$

Substitute \frac{40}{3}x for y in the equation x=a-y.

$$x=a-\frac{40}{3}x$$

Solve x=a-\frac{40}{3}x for x.

$$x=\frac{3}{43}a$$

Substitute \frac{3}{43}a for x in the equation y=\frac{40}{3}x.

$$y=\frac{40}{3}\times \frac{3}{43}a$$

Calculate y from y=\frac{40}{3}\times \frac{3}{43}a.

$$y=\frac{40}{43}a$$

Substitute \frac{40}{43}a for y and \frac{3}{43}a for x in the equation z=x+y.

$$z=\frac{3}{43}a+\frac{40}{43}a$$

Calculate z from z=\frac{3}{43}a+\frac{40}{43}a.

$$z=a$$

The system is now solved.

$$x=\frac{3}{43}a y=\frac{40}{43}a z=a$$