Solve {l}{0=80x+30y+8}{y=70z+7}{z=8x+3}

$$\left. \begin{array} { l } { 0 = 80 x + 30 y + 8 } \\ { y = 70 z + 7 } \\ { z = 8 x + 3 } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{3259}{8440}\approx -0.386137441$$

$$y=\frac{161}{211}\approx 0.763033175$$

$$z=-\frac{94}{1055}\approx -0.089099526$$

Short Steps Using Substitution

Reorder the equations.

$$y=70z+7 0=80x+30y+8 z=8x+3$$

Substitute 7+70z for y in the equation 0=80x+30y+8.

$$0=80x+30\left(7+70z\right)+8$$

Solve 0=80x+30\left(7+70z\right)+8 for x.

$$x=-\frac{109}{40}-\frac{105}{4}z$$

Substitute -\frac{109}{40}-\frac{105}{4}z for x in the equation z=8x+3.

$$z=8\left(-\frac{109}{40}-\frac{105}{4}z\right)+3$$

Solve z=8\left(-\frac{109}{40}-\frac{105}{4}z\right)+3 for z.

$$z=-\frac{94}{1055}$$

Substitute -\frac{94}{1055} for z in the equation x=-\frac{109}{40}-\frac{105}{4}z.

$$x=-\frac{109}{40}-\frac{105}{4}\left(-\frac{94}{1055}\right)$$

Calculate x from x=-\frac{109}{40}-\frac{105}{4}\left(-\frac{94}{1055}\right).

$$x=-\frac{3259}{8440}$$

Substitute -\frac{94}{1055} for z in the equation y=70z+7.

$$y=70\left(-\frac{94}{1055}\right)+7$$

Calculate y from y=70\left(-\frac{94}{1055}\right)+7.

$$y=\frac{161}{211}$$

The system is now solved.

$$x=-\frac{3259}{8440} y=\frac{161}{211} z=-\frac{94}{1055}$$