Solve {l}{25{(x+2)}=0*5{(2x-1)}}{y=0}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 25 {(x + 2)} = 0 \cdot 5 {(2 x - 1)} }\\ { y = 0 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=0$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 25 by x+2.

$$25x+50=0\times 5\left(2x-1\right)$$

Multiply 0 and 5 to get 0.

$$25x+50=0\left(2x-1\right)$$

Anything times zero gives zero.

$$25x+50=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$25x=-50$$

Divide both sides by 25.

$$x=\frac{-50}{25}$$

Divide -50 by 25 to get -2.

$$x=-2$$

The system is now solved.

$$x=-2 y=0 z=0$$