Solve {l}{3y+x=z}{z=N+1}{y-x=5}

$$\left. \begin{array} { l } { 3 y + x = z } \\ { z = N + 1 } \\ { y - x = 5 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{N-14}{4}$$

$$y=\frac{N+6}{4}$$

$$z=N+1$$

Short Steps Using Substitution

Solve 3y+x=z for z.

$$z=3y+x$$

Substitute 3y+x for z in the equation z=N+1.

$$3y+x=N+1$$

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

$$y=\frac{1}{3}-\frac{1}{3}x+\frac{1}{3}N x=y-5$$

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

$$x=\frac{1}{3}-\frac{1}{3}x+\frac{1}{3}N-5$$

Solve x=\frac{1}{3}-\frac{1}{3}x+\frac{1}{3}N-5 for x.

$$x=-\frac{7}{2}+\frac{1}{4}N$$

Substitute -\frac{7}{2}+\frac{1}{4}N for x in the equation y=\frac{1}{3}-\frac{1}{3}x+\frac{1}{3}N.

$$y=\frac{1}{3}-\frac{1}{3}\left(-\frac{7}{2}+\frac{1}{4}N\right)+\frac{1}{3}N$$

Calculate y from y=\frac{1}{3}-\frac{1}{3}\left(-\frac{7}{2}+\frac{1}{4}N\right)+\frac{1}{3}N.

$$y=\frac{3}{2}+\frac{1}{4}N$$

Substitute \frac{3}{2}+\frac{1}{4}N for y and -\frac{7}{2}+\frac{1}{4}N for x in the equation z=3y+x.

$$z=3\left(\frac{3}{2}+\frac{1}{4}N\right)-\frac{7}{2}+\frac{1}{4}N$$

Calculate z from z=3\left(\frac{3}{2}+\frac{1}{4}N\right)-\frac{7}{2}+\frac{1}{4}N.

$$z=1+N$$

The system is now solved.

$$x=-\frac{7}{2}+\frac{1}{4}N y=\frac{3}{2}+\frac{1}{4}N z=1+N$$