Solve {l}{w+2x+4y-z=5}{2x-4y+z=5}{w-4x+y=5}

$$\left. \begin{array}{l}{ w + 2 x + 4 y - z = 5 }\\{ 2 x - 4 y + z = 5 }\\{ w - 4 x + y = 5 }\end{array} \right.$$

Solve for x, y, z

$$x=-\frac{w}{4}+\frac{5}{2}$$

$$y=15-2w$$

$$z=-\frac{15w}{2}+60$$

Short Steps Using Substitution

Solve w+2x+4y-z=5 for z.

$$z=w+2x+4y-5$$

Substitute w+2x+4y-5 for z in the equation 2x-4y+z=5.

$$2x-4y+w+2x+4y-5=5$$

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

$$x=\frac{5}{2}-\frac{1}{4}w y=-w+4x+5$$

Substitute \frac{5}{2}-\frac{1}{4}w for x in the equation y=-w+4x+5.

$$y=-w+4\left(\frac{5}{2}-\frac{1}{4}w\right)+5$$

Calculate y from y=-w+4\left(\frac{5}{2}-\frac{1}{4}w\right)+5.

$$y=15-2w$$

Substitute \frac{5}{2}-\frac{1}{4}w for x and 15-2w for y in the equation z=w+2x+4y-5.

$$z=w+2\left(\frac{5}{2}-\frac{1}{4}w\right)+4\left(15-2w\right)-5$$

Calculate z from z=w+2\left(\frac{5}{2}-\frac{1}{4}w\right)+4\left(15-2w\right)-5.

$$z=60-\frac{15}{2}w$$

The system is now solved.

$$x=\frac{5}{2}-\frac{1}{4}w y=15-2w z=60-\frac{15}{2}w$$