Solve {l}{d+b=3}{d-c=8}{b-2c=4}

$$\left\{ \begin{array} { l } { d + b = 3 } \\ { d - c = 8 } \\ { b - 2 c = 4 } \end{array} \right.$$

Solve for d, b, c

$$d=5$$

$$b=-2$$

$$c=-3$$

Short Steps Using Substitution

Solve d+b=3 for d.

$$d=-b+3$$

Substitute -b+3 for d in the equation d-c=8.

$$-b+3-c=8$$

Solve the second equation for b and the third equation for c.

$$b=-5-c c=\frac{1}{2}b-2$$

Substitute -5-c for b in the equation c=\frac{1}{2}b-2.

$$c=\frac{1}{2}\left(-5-c\right)-2$$

Solve c=\frac{1}{2}\left(-5-c\right)-2 for c.

$$c=-3$$

Substitute -3 for c in the equation b=-5-c.

$$b=-5-\left(-3\right)$$

Calculate b from b=-5-\left(-3\right).

$$b=-2$$

Substitute -2 for b in the equation d=-b+3.

$$d=-\left(-2\right)+3$$

Calculate d from d=-\left(-2\right)+3.

$$d=5$$

The system is now solved.

$$d=5 b=-2 c=-3$$