Solve {l}{2x+1=3x}{y=x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 2 x + 1 = 3 x }\\ { y = x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=1$$

Solution Steps

Consider the first equation. Subtract 3x from both sides.

$$2x+1-3x=0$$

Combine 2x and -3x to get -x.

$$-x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-x=-1$$

Divide both sides by -1.

$$x=\frac{-1}{-1}$$

Divide -1 by -1 to get 1.

$$x=1$$

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

$$y=1$$

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

$$z=1$$

The system is now solved.

$$x=1 y=1 z=1$$