Solve {l}{x/{(frac{2*{(2)}+1}{2})}=0}{y=6/{(frac{1*{(5)}+1}{5})}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { x / {(\frac{2 \cdot {(2)} + 1}{2})} = 0 }\\ { y = 6 / {(\frac{1 \cdot {(5)} + 1}{5})} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=5$$

Solution Steps

Consider the first equation. Divide x by \frac{2\times 2+1}{2} by multiplying x by the reciprocal of \frac{2\times 2+1}{2}.

$$\frac{x\times 2}{2\times 2+1}=0$$

Multiply 2 and 2 to get 4.

$$\frac{x\times 2}{4+1}=0$$

Add 4 and 1 to get 5.

$$\frac{x\times 2}{5}=0$$

Multiply both sides by 5. Anything times zero gives zero.

$$x\times 2=0$$

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$x=0$$

Consider the second equation. Divide 6 by \frac{1\times 5+1}{5} by multiplying 6 by the reciprocal of \frac{1\times 5+1}{5}.

$$y=\frac{6\times 5}{1\times 5+1}$$

Multiply 6 and 5 to get 30.

$$y=\frac{30}{1\times 5+1}$$

Multiply 1 and 5 to get 5.

$$y=\frac{30}{5+1}$$

Add 5 and 1 to get 6.

$$y=\frac{30}{6}$$

Divide 30 by 6 to get 5.

$$y=5$$

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

$$z=5$$

The system is now solved.

$$x=0 y=5 z=5$$