Solve {l}{x+y=48}{6=y}{z=x+0}{a=z}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

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

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

$$d=42$$

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.

$$x+6=48$$

Subtract 6 from both sides.

$$x=48-6$$

Subtract 6 from 48 to get 42.

$$x=42$$

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

$$z=42+0$$

Add 42 and 0 to get 42.

$$z=42$$

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

$$a=42$$

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

$$b=42$$

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

$$c=42$$

Consider the equation (7). Insert the known values of variables into the equation.

$$d=42$$

The system is now solved.

$$x=42 y=6 z=42 a=42 b=42 c=42 d=42$$