Solve {l}{x/12+8={(x+2)}*4}{y=0}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { \frac{x}{12} + 8 = {(x + 2)} \cdot 4 }\\ { y = 0 }\\ { z = y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for x, y, z, a, b, c

$$c=0$$

Solution Steps

Consider the first equation. Multiply both sides of the equation by 12.

$$x+96=12\left(x+2\right)\times 4$$

Multiply 12 and 4 to get 48.

$$x+96=48\left(x+2\right)$$

Use the distributive property to multiply 48 by x+2.

$$x+96=48x+96$$

Subtract 48x from both sides.

$$x+96-48x=96$$

Combine x and -48x to get -47x.

$$-47x+96=96$$

Subtract 96 from both sides.

$$-47x=96-96$$

Subtract 96 from 96 to get 0.

$$-47x=0$$

Divide both sides by -47. Zero divided by any non-zero number gives zero.

$$x=0$$

The system is now solved.

$$x=0 y=0 z=0 a=0 b=0 c=0$$