Solve {r}{x-y=1}{4x-8z=16}{4y+z=12}

$$\left\{ \begin{array} { r } { x - y = 1 } \\ { 4 x - 8 z = 16 } \\ { 4 y + z = 12 } \end{array} \right.$$

Solve for x, y, z

$$x=4$$

$$y=3$$

$$z=0$$

Short Steps Using Substitution

Solve x-y=1 for x.

$$x=y+1$$

Substitute y+1 for x in the equation 4x-8z=16.

$$4\left(y+1\right)-8z=16$$

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

$$y=3+2z z=12-4y$$

Substitute 3+2z for y in the equation z=12-4y.

$$z=12-4\left(3+2z\right)$$

Solve z=12-4\left(3+2z\right) for z.

$$z=0$$

Substitute 0 for z in the equation y=3+2z.

$$y=3+2\times 0$$

Calculate y from y=3+2\times 0.

$$y=3$$

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

$$x=3+1$$

Calculate x from x=3+1.

$$x=4$$

The system is now solved.

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