Solve {l}{9-2{(x+4)}-10{(25-x+4)}=5-3x-4{(x+1)}}{y=7x13*7x17*3x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 9 - 2 {(x + 4)} - 10 {(25 - x + 4)} = 5 - 3 x - 4 {(x + 1)} }\\ { y = 7 x 13 \cdot 7 x 17 \cdot 3 x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=2842x_{13}x_{17}$$

Short Steps Using Substitution

Reorder the equations.

$$y=7x_{13}\times 7x_{17}\times 3x 9-2\left(x+4\right)-10\left(25-x+4\right)=5-3x-4\left(x+1\right) z=y$$

Calculate y from y=7x_{13}\times 7x_{17}\times 3x.

$$y=147x_{17}x_{13}x$$

Substitute 147x_{17}x_{13}x for y in the equation z=y.

$$z=147x_{17}x_{13}x$$

Solve 9-2\left(x+4\right)-10\left(25-x+4\right)=5-3x-4\left(x+1\right) for x.

$$x=\frac{58}{3}$$

Substitute \frac{58}{3} for x in the equation z=147x_{17}x_{13}x.

$$z=147x_{17}x_{13}\times \frac{58}{3}$$

Calculate z from z=147x_{17}x_{13}\times \frac{58}{3}.

$$z=2842x_{17}x_{13}$$

Substitute \frac{58}{3} for x in the equation y=147x_{17}x_{13}x.

$$y=147x_{17}x_{13}\times \frac{58}{3}$$

Calculate y from y=147x_{17}x_{13}\times \frac{58}{3}.

$$y=2842x_{17}x_{13}$$

The system is now solved.

$$x=\frac{58}{3} y=2842x_{17}x_{13} z=2842x_{17}x_{13}$$