Solve {l}{2x+4z+3=0}{3x+2z+4=9}{y-2z=4}

$$\left. \begin{array} { l } { 2 x + 4 z + 3 = 0 } \\ { 3 x + 2 z + 4 = 9 } \\ { y - 2 z = 4 } \end{array} \right.$$

Solve for x, z, y

$$x = \frac{13}{4} = 3\frac{1}{4} = 3.25$$

$$y=-\frac{3}{4}=-0.75$$

$$z = -\frac{19}{8} = -2\frac{3}{8} = -2.375$$

Short Steps Using Substitution

Solve 2x+4z+3=0 for x.

$$x=-2z-\frac{3}{2}$$

Substitute -2z-\frac{3}{2} for x in the equation 3x+2z+4=9.

$$3\left(-2z-\frac{3}{2}\right)+2z+4=9$$

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

$$z=-\frac{19}{8} y=4+2z$$

Substitute -\frac{19}{8} for z in the equation y=4+2z.

$$y=4+2\left(-\frac{19}{8}\right)$$

Calculate y from y=4+2\left(-\frac{19}{8}\right).

$$y=-\frac{3}{4}$$

Substitute -\frac{19}{8} for z in the equation x=-2z-\frac{3}{2}.

$$x=-2\left(-\frac{19}{8}\right)-\frac{3}{2}$$

Calculate x from x=-2\left(-\frac{19}{8}\right)-\frac{3}{2}.

$$x=\frac{13}{4}$$

The system is now solved.

$$x=\frac{13}{4} z=-\frac{19}{8} y=-\frac{3}{4}$$