Solve {l}{y=z+x}{z-x+4y=0}{y=3}

$$\left. \begin{array} { l } { y = z + x } \\ { z - x + 4 y = 0 } \\ { y = 3 } \end{array} \right.$$

Solve for y, z, x

$$x = \frac{15}{2} = 7\frac{1}{2} = 7.5$$

$$y=3$$

$$z = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

Short Steps Using Substitution

Reorder the equations.

$$y=3 z-x+4y=0 y=z+x$$

Substitute 3 for y in the second and third equation.

$$z-x+4\times 3=0 3=z+x$$

Solve these equations for z and x respectively.

$$z=x-12 x=3-z$$

Substitute x-12 for z in the equation x=3-z.

$$x=3-\left(x-12\right)$$

Solve x=3-\left(x-12\right) for x.

$$x=\frac{15}{2}$$

Substitute \frac{15}{2} for x in the equation z=x-12.

$$z=\frac{15}{2}-12$$

Calculate z from z=\frac{15}{2}-12.

$$z=-\frac{9}{2}$$

The system is now solved.

$$y=3 z=-\frac{9}{2} x=\frac{15}{2}$$