Solve {l}{2-2{(3x-y)}=10{(4-y)}}{z=4{(y-x)}}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 2 - 2 {(3 x - y)} = 10 {(4 - y)} }\\ { z = 4 {(y - x)} }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{a}{2}+\frac{19}{3}$$

$$y=-\frac{a}{4}+\frac{19}{3}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=4\left(y-x\right) 2-2\left(3x-y\right)=10\left(4-y\right) a=z$$

Calculate z from z=4\left(y-x\right).

$$z=4y-4x$$

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

$$a=4y-4x$$

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

$$y=\frac{19}{6}+\frac{1}{2}x x=-\frac{1}{4}a+y$$

Substitute \frac{19}{6}+\frac{1}{2}x for y in the equation x=-\frac{1}{4}a+y.

$$x=-\frac{1}{4}a+\frac{19}{6}+\frac{1}{2}x$$

Solve x=-\frac{1}{4}a+\frac{19}{6}+\frac{1}{2}x for x.

$$x=\frac{19}{3}-\frac{1}{2}a$$

Substitute \frac{19}{3}-\frac{1}{2}a for x in the equation y=\frac{19}{6}+\frac{1}{2}x.

$$y=\frac{19}{6}+\frac{1}{2}\left(\frac{19}{3}-\frac{1}{2}a\right)$$

Calculate y from y=\frac{19}{6}+\frac{1}{2}\left(\frac{19}{3}-\frac{1}{2}a\right).

$$y=\frac{19}{3}-\frac{1}{4}a$$

Substitute \frac{19}{3}-\frac{1}{4}a for y and \frac{19}{3}-\frac{1}{2}a for x in the equation z=4y-4x.

$$z=4\left(\frac{19}{3}-\frac{1}{4}a\right)-4\left(\frac{19}{3}-\frac{1}{2}a\right)$$

Calculate z from z=4\left(\frac{19}{3}-\frac{1}{4}a\right)-4\left(\frac{19}{3}-\frac{1}{2}a\right).

$$z=a$$

The system is now solved.

$$x=\frac{19}{3}-\frac{1}{2}a y=\frac{19}{3}-\frac{1}{4}a z=a$$