Solve {l}{(2)^2+x=1}{y=2x(1)^2}{text{Solvefor}ztext{where}}{z=y}

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

Solve for x, y, z

$$z=-6$$

Solution Steps

Consider the first equation. Calculate 2 to the power of 2 and get 4.

$$4+x=1$$

Subtract 4 from both sides.

$$x=1-4$$

Subtract 4 from 1 to get -3.

$$x=-3$$

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

$$y=2\left(-3\right)\times 1^{2}$$

Multiply 2 and -3 to get -6.

$$y=-6\times 1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$y=-6$$

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

$$z=-6$$

The system is now solved.

$$x=-3 y=-6 z=-6$$