Solve {c}{x*y=z}{(x+5)*(y-200)=z}{(x-4)(y-232)=z}

$$\left\{ \begin{array} { c } { x \cdot y = z } \\ { ( x + 5 ) \cdot ( y - 200 ) = z } \\ { ( x - 4 ) ( y - 232 ) = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{16}{49}\approx 0.326530612$$

$$y = \frac{10440}{49} = 213\frac{3}{49} \approx 213.06122449$$

$$z = \frac{167040}{2401} = 69\frac{1371}{2401} \approx 69.571012078$$

Short Steps Using Substitution

Solve xy=z for z.

$$z=xy$$

Substitute xy for z in the second and third equation.

$$\left(x+5\right)\left(y-200\right)=xy \left(x-4\right)\left(y-232\right)=xy$$

Solve these equations for y and x respectively.

$$y=200+40x x=4-\frac{1}{58}y$$

Substitute 200+40x for y in the equation x=4-\frac{1}{58}y.

$$x=4-\frac{1}{58}\left(200+40x\right)$$

Solve x=4-\frac{1}{58}\left(200+40x\right) for x.

$$x=\frac{16}{49}$$

Substitute \frac{16}{49} for x in the equation y=200+40x.

$$y=200+40\times \frac{16}{49}$$

Calculate y from y=200+40\times \frac{16}{49}.

$$y=\frac{10440}{49}$$

Substitute \frac{10440}{49} for y and \frac{16}{49} for x in the equation z=xy.

$$z=\frac{16}{49}\times \frac{10440}{49}$$

Calculate z from z=\frac{16}{49}\times \frac{10440}{49}.

$$z=\frac{167040}{2401}$$

The system is now solved.

$$x=\frac{16}{49} y=\frac{10440}{49} z=\frac{167040}{2401}$$