Solve {l}{{({(7y+13-5x)}/4)}+y=2x-{(3y+2{(x-8)})}/3}{z=x-4{(3-2y)}}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { {({(7 y + 13 - 5 x)} / 4)} + y = 2 x - {(3 y + 2 {(x - 8)})} / 3 }\\ { z = x - 4 {(3 - 2 y)} }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{45a+340}{293}$$

$$y=\frac{31a+397}{293}$$

$$z=a$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$39+33y-15x=64+16x-12y z=x-4\left(3-2y\right) a=z$$

Reorder the equations.

$$z=x-4\left(3-2y\right) 39+33y-15x=64+16x-12y a=z$$

Calculate z from z=x-4\left(3-2y\right).

$$z=x-12+8y$$

Substitute x-12+8y for z in the equation a=z.

$$a=x-12+8y$$

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

$$y=\frac{5}{9}+\frac{31}{45}x x=12+a-8y$$

Substitute \frac{5}{9}+\frac{31}{45}x for y in the equation x=12+a-8y.

$$x=12+a-8\left(\frac{5}{9}+\frac{31}{45}x\right)$$

Solve x=12+a-8\left(\frac{5}{9}+\frac{31}{45}x\right) for x.

$$x=\frac{340}{293}+\frac{45}{293}a$$

Substitute \frac{340}{293}+\frac{45}{293}a for x in the equation y=\frac{5}{9}+\frac{31}{45}x.

$$y=\frac{5}{9}+\frac{31}{45}\left(\frac{340}{293}+\frac{45}{293}a\right)$$

Calculate y from y=\frac{5}{9}+\frac{31}{45}\left(\frac{340}{293}+\frac{45}{293}a\right).

$$y=\frac{397}{293}+\frac{31}{293}a$$

Substitute \frac{397}{293}+\frac{31}{293}a for y and \frac{340}{293}+\frac{45}{293}a for x in the equation z=x-12+8y.

$$z=\frac{340}{293}+\frac{45}{293}a-12+8\left(\frac{397}{293}+\frac{31}{293}a\right)$$

Calculate z from z=\frac{340}{293}+\frac{45}{293}a-12+8\left(\frac{397}{293}+\frac{31}{293}a\right).

$$z=a$$

The system is now solved.

$$x=\frac{340}{293}+\frac{45}{293}a y=\frac{397}{293}+\frac{31}{293}a z=a$$