Solve {l}{2+y=1*5*y}{z=1}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 2 + y = 1 \cdot 5 \cdot y }\\ { z = 1 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for y, z, a

$$a=1$$

Solution Steps

Consider the first equation. Multiply 1 and 5 to get 5.

$$2+y=5y$$

Subtract 5y from both sides.

$$2+y-5y=0$$

Combine y and -5y to get -4y.

$$2-4y=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$-4y=-2$$

Divide both sides by -4.

$$y=\frac{-2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out -2.

$$y=\frac{1}{2}$$

The system is now solved.

$$y=\frac{1}{2} z=1 a=1$$