Solve {l}{6x+3y=7}{4y+6z=6}{z+y=1}

$$\left. \begin{array} { l } { 6 x + 3 y = 7 } \\ { 4 y + 6 z = 6 } \\ { z + y = 1 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

$$y=0$$

$$z=1$$

Short Steps Using Substitution

Reorder the equations.

$$z+y=1 4y+6z=6 6x+3y=7$$

Solve z+y=1 for y.

$$y=1-z$$

Substitute 1-z for y in the second and third equation.

$$4\left(1-z\right)+6z=6 6x+3\left(1-z\right)=7$$

Solve these equations for z and x respectively.

$$z=1 x=\frac{2}{3}+\frac{1}{2}z$$

Substitute 1 for z in the equation x=\frac{2}{3}+\frac{1}{2}z.

$$x=\frac{2}{3}+\frac{1}{2}\times 1$$

Calculate x from x=\frac{2}{3}+\frac{1}{2}\times 1.

$$x=\frac{7}{6}$$

Substitute 1 for z in the equation y=1-z.

$$y=1-1$$

Calculate y from y=1-1.

$$y=0$$

The system is now solved.

$$x=\frac{7}{6} y=0 z=1$$