Solve {l}{y-4=0}{z=2x+ky}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { y - 4 = 0 }\\ { z = 2 x + k y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a-4k}{2}$$

$$y=4$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=2x+ky y-4=0 a=z$$

Substitute 2x+ky for z in the equation a=z.

$$a=2x+ky$$

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

$$y=4 x=-\frac{1}{2}ky+\frac{1}{2}a$$

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

$$x=-\frac{1}{2}k\times 4+\frac{1}{2}a$$

Calculate x from x=-\frac{1}{2}k\times 4+\frac{1}{2}a.

$$x=\frac{1}{2}a-2k$$

Substitute 4 for y and \frac{1}{2}a-2k for x in the equation z=2x+ky.

$$z=2\left(\frac{1}{2}a-2k\right)+k\times 4$$

Calculate z from z=2\left(\frac{1}{2}a-2k\right)+k\times 4.

$$z=a$$

The system is now solved.

$$x=\frac{1}{2}a-2k y=4 z=a$$