Solve {l}{5{(3x-1)}+132*4={(2*8-4x)}*0*5}{y=12}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 5 {(3 x - 1)} + 132 \cdot 4 = {(2 \cdot 8 - 4 x)} \cdot 0 \cdot 5 }\\ { y = 12 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=12$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 5 by 3x-1.

$$15x-5+132\times 4=\left(2\times 8-4x\right)\times 0\times 5$$

Multiply 132 and 4 to get 528.

$$15x-5+528=\left(2\times 8-4x\right)\times 0\times 5$$

Add -5 and 528 to get 523.

$$15x+523=\left(2\times 8-4x\right)\times 0\times 5$$

Multiply 2 and 8 to get 16.

$$15x+523=\left(16-4x\right)\times 0\times 5$$

Multiply 0 and 5 to get 0.

$$15x+523=\left(16-4x\right)\times 0$$

Anything times zero gives zero.

$$15x+523=0$$

Subtract 523 from both sides. Anything subtracted from zero gives its negation.

$$15x=-523$$

Divide both sides by 15.

$$x=-\frac{523}{15}$$

The system is now solved.

$$x=-\frac{523}{15} y=12 z=12$$