Solve {l}{4y+2z-t=0}{y+z+t=-1}{2z+3t=-4}

$$\left\{ \begin{array} { l } { 4 y + 2 z - t = 0 } \\ { y + z + t = - 1 } \\ { 2 z + 3 t = - 4 } \end{array} \right.$$

Solve for y, z, t

$$y=1$$

$$z=-2$$

$$t=0$$

Short Steps Using Substitution

Solve 4y+2z-t=0 for t.

$$t=4y+2z$$

Substitute 4y+2z for t in the second and third equation.

$$y+z+4y+2z=-1 2z+3\left(4y+2z\right)=-4$$

Solve these equations for z and y respectively.

$$z=-\frac{5}{3}y-\frac{1}{3} y=-\frac{1}{3}-\frac{2}{3}z$$

Substitute -\frac{5}{3}y-\frac{1}{3} for z in the equation y=-\frac{1}{3}-\frac{2}{3}z.

$$y=-\frac{1}{3}-\frac{2}{3}\left(-\frac{5}{3}y-\frac{1}{3}\right)$$

Solve y=-\frac{1}{3}-\frac{2}{3}\left(-\frac{5}{3}y-\frac{1}{3}\right) for y.

$$y=1$$

Substitute 1 for y in the equation z=-\frac{5}{3}y-\frac{1}{3}.

$$z=-\frac{5}{3}-\frac{1}{3}$$

Calculate z from z=-\frac{5}{3}-\frac{1}{3}.

$$z=-2$$

Substitute -2 for z and 1 for y in the equation t=4y+2z.

$$t=4\times 1+2\left(-2\right)$$

Calculate t from t=4\times 1+2\left(-2\right).

$$t=0$$

The system is now solved.

$$y=1 z=-2 t=0$$