Solve {l}{4x+y=3}{z=5x+7}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 4 x + y = 3 }\\ { z = 5 x + 7 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a-7}{5}$$

$$y=\frac{43-4a}{5}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=5x+7 4x+y=3 a=z$$

Substitute 5x+7 for z in the equation a=z.

$$a=5x+7$$

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

$$y=-4x+3 x=-\frac{7}{5}+\frac{1}{5}a$$

Substitute -\frac{7}{5}+\frac{1}{5}a for x in the equation y=-4x+3.

$$y=-4\left(-\frac{7}{5}+\frac{1}{5}a\right)+3$$

Calculate y from y=-4\left(-\frac{7}{5}+\frac{1}{5}a\right)+3.

$$y=\frac{43}{5}-\frac{4}{5}a$$

Substitute -\frac{7}{5}+\frac{1}{5}a for x in the equation z=5x+7.

$$z=5\left(-\frac{7}{5}+\frac{1}{5}a\right)+7$$

Calculate z from z=5\left(-\frac{7}{5}+\frac{1}{5}a\right)+7.

$$z=a$$

The system is now solved.

$$x=-\frac{7}{5}+\frac{1}{5}a y=\frac{43}{5}-\frac{4}{5}a z=a$$