Solve {l}{{(2)}-x+y=48}{6=y}{z=x+0}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { {(2)} - x + y = 48 }\\ { 6 = y }\\ { z = x + 0 }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for x, y, z, a, b, c

$$c=-40$$

Solution Steps

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$y=6$$

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

$$2-x+6=48$$

Add 2 and 6 to get 8.

$$8-x=48$$

Subtract 8 from both sides.

$$-x=48-8$$

Subtract 8 from 48 to get 40.

$$-x=40$$

Divide both sides by -1.

$$x=-40$$

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

$$z=-40+0$$

Add -40 and 0 to get -40.

$$z=-40$$

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

$$a=-40$$

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

$$b=-40$$

Consider the equation (6). Insert the known values of variables into the equation.

$$c=-40$$

The system is now solved.

$$x=-40 y=6 z=-40 a=-40 b=-40 c=-40$$