Solve {l}{x+y=4}{z=4x-10}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y = 4 }\\ { z = 4 x - 10 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a}{4}+\frac{5}{2}$$

$$y=-\frac{a}{4}+\frac{3}{2}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=4x-10 x+y=4 a=z$$

Substitute -10+4x for z in the equation a=z.

$$a=-10+4x$$

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

$$y=-x+4 x=\frac{5}{2}+\frac{1}{4}a$$

Substitute \frac{5}{2}+\frac{1}{4}a for x in the equation y=-x+4.

$$y=-\left(\frac{5}{2}+\frac{1}{4}a\right)+4$$

Calculate y from y=-\left(\frac{5}{2}+\frac{1}{4}a\right)+4.

$$y=\frac{3}{2}-\frac{1}{4}a$$

Substitute \frac{5}{2}+\frac{1}{4}a for x in the equation z=4x-10.

$$z=4\left(\frac{5}{2}+\frac{1}{4}a\right)-10$$

Calculate z from z=4\left(\frac{5}{2}+\frac{1}{4}a\right)-10.

$$z=a$$

The system is now solved.

$$x=\frac{5}{2}+\frac{1}{4}a y=\frac{3}{2}-\frac{1}{4}a z=a$$