Solve {l}{-55+B-5C+D=0}{-88+4B-10C+D=0}{-88+5B-8C+D=0}

$$\left\{ \begin{array} { l } { - 55 + B - 5 C + D = 0 } \\ { - 88 + 4 B - 10 C + D = 0 } \\ { - 88 + 5 B - 8 C + D = 0 } \end{array} \right.$$

Solve for B, C, D

$$B=6$$

$$C=-3$$

$$D=34$$

Short Steps Using Substitution

Solve -55+B-5C+D=0 for B.

$$B=55+5C-D$$

Substitute 55+5C-D for B in the second and third equation.

$$-88+4\left(55+5C-D\right)-10C+D=0 -88+5\left(55+5C-D\right)-8C+D=0$$

Solve these equations for C and D respectively.

$$C=-\frac{66}{5}+\frac{3}{10}D D=\frac{187}{4}+\frac{17}{4}C$$

Substitute -\frac{66}{5}+\frac{3}{10}D for C in the equation D=\frac{187}{4}+\frac{17}{4}C.

$$D=\frac{187}{4}+\frac{17}{4}\left(-\frac{66}{5}+\frac{3}{10}D\right)$$

Solve D=\frac{187}{4}+\frac{17}{4}\left(-\frac{66}{5}+\frac{3}{10}D\right) for D.

$$D=34$$

Substitute 34 for D in the equation C=-\frac{66}{5}+\frac{3}{10}D.

$$C=-\frac{66}{5}+\frac{3}{10}\times 34$$

Calculate C from C=-\frac{66}{5}+\frac{3}{10}\times 34.

$$C=-3$$

Substitute -3 for C and 34 for D in the equation B=55+5C-D.

$$B=55+5\left(-3\right)-34$$

Calculate B from B=55+5\left(-3\right)-34.

$$B=6$$

The system is now solved.

$$B=6 C=-3 D=34$$