Solve {l}{1+2x+x+x+16=220}{y={(2x+x+x+16)}*0}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 1 + 2 x + x + x + 16 = 220 }\\ { y = {(2 x + x + x + 16)} \cdot 0 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=0$$

Solution Steps

Consider the first equation. Combine 2x and x to get 3x.

$$1+3x+x+16=220$$

Combine 3x and x to get 4x.

$$1+4x+16=220$$

Add 1 and 16 to get 17.

$$17+4x=220$$

Subtract 17 from both sides.

$$4x=220-17$$

Subtract 17 from 220 to get 203.

$$4x=203$$

Divide both sides by 4.

$$x=\frac{203}{4}$$

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

$$y=\left(2\times \frac{203}{4}+\frac{203}{4}+\frac{203}{4}+16\right)\times 0$$

Multiply 2 and \frac{203}{4} to get \frac{203}{2}.

$$y=\left(\frac{203}{2}+\frac{203}{4}+\frac{203}{4}+16\right)\times 0$$

Add \frac{203}{2} and \frac{203}{4} to get \frac{609}{4}.

$$y=\left(\frac{609}{4}+\frac{203}{4}+16\right)\times 0$$

Add \frac{609}{4} and \frac{203}{4} to get 203.

$$y=\left(203+16\right)\times 0$$

Add 203 and 16 to get 219.

$$y=219\times 0$$

Multiply 219 and 0 to get 0.

$$y=0$$

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

$$z=0$$

The system is now solved.

$$x=\frac{203}{4} y=0 z=0$$