Solve {l}{27/8+3/2b+c=-1/4}{27/4+b=0}

$$\left. \begin{array} { l } { \frac { 27 } { 8 } + \frac { 3 } { 2 } b + c = - \frac { 1 } { 4 } } \\ { \frac { 27 } { 4 } + b = 0 } \end{array} \right.$$

Solve for b, c

$$b = -\frac{27}{4} = -6\frac{3}{4} = -6.75$$

$$c = \frac{13}{2} = 6\frac{1}{2} = 6.5$$

Solution Steps

Consider the second equation. Subtract \frac{27}{4} from both sides. Anything subtracted from zero gives its negation.

$$b=-\frac{27}{4}$$

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

$$\frac{27}{8}+\frac{3}{2}\left(-\frac{27}{4}\right)+c=-\frac{1}{4}$$

Multiply \frac{3}{2} and -\frac{27}{4} to get -\frac{81}{8}.

$$\frac{27}{8}-\frac{81}{8}+c=-\frac{1}{4}$$

Subtract \frac{81}{8} from \frac{27}{8} to get -\frac{27}{4}.

$$-\frac{27}{4}+c=-\frac{1}{4}$$

Add \frac{27}{4} to both sides.

$$c=-\frac{1}{4}+\frac{27}{4}$$

Add -\frac{1}{4} and \frac{27}{4} to get \frac{13}{2}.

$$c=\frac{13}{2}$$

The system is now solved.

$$b=-\frac{27}{4} c=\frac{13}{2}$$