Solve {l}{8/x=4/x+4}{y=x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { \frac{8}{x} = \frac{4}{x + 4} }\\ { y = x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=-8$$

Solution Steps

Consider the first equation. Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x,x+4.

$$\left(x+4\right)\times 8=x\times 4$$

Use the distributive property to multiply x+4 by 8.

$$8x+32=x\times 4$$

Subtract x\times 4 from both sides.

$$8x+32-x\times 4=0$$

Combine 8x and -x\times 4 to get 4x.

$$4x+32=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$4x=-32$$

Divide both sides by 4.

$$x=\frac{-32}{4}$$

Divide -32 by 4 to get -8.

$$x=-8$$

Consider the second equation. Insert the known values of variables into the equation.

$$y=-8$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=-8$$

The system is now solved.

$$x=-8 y=-8 z=-8$$