Solve {l}{y=2{(3y-1/2)}+1}{z=y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { y = 2 {(\frac{3 y - 1}{2})} + 1 }\\ { z = y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for y, z, a

$$a=0$$

Solution Steps

Consider the first equation. Express 2\times \frac{3y-1}{2} as a single fraction.

$$y=\frac{2\left(3y-1\right)}{2}+1$$

Cancel out 2 and 2.

$$y=3y-1+1$$

Add -1 and 1 to get 0.

$$y=3y$$

Subtract 3y from both sides.

$$y-3y=0$$

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

$$-2y=0$$

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

$$y=0$$

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

$$z=0$$

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

$$a=0$$

The system is now solved.

$$y=0 z=0 a=0$$