Solve {l}{x=260-150+0*1/2}{y={(150+x)}}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { x = 260 - 150 + \frac{0 \cdot 1}{2} }\\ { y = {(150 + x)} }\\ { z = y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

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

$$c=260$$

Solution Steps

Consider the first equation. Subtract 150 from 260 to get 110.

$$x=110+\frac{0\times 1}{2}$$

Multiply 0 and 1 to get 0.

$$x=110+\frac{0}{2}$$

Zero divided by any non-zero number gives zero.

$$x=110+0$$

Add 110 and 0 to get 110.

$$x=110$$

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

$$y=150+110$$

Add 150 and 110 to get 260.

$$y=260$$

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

$$z=260$$

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

$$a=260$$

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

$$b=260$$

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

$$c=260$$

The system is now solved.

$$x=110 y=260 z=260 a=260 b=260 c=260$$