Solve {l}{7/8b+1/4=1}{c=2}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { \frac{7}{8} b + \frac{1}{4} = 1 }\\ { c = 2 }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for b, c, d

$$d=2$$

Solution Steps

Consider the first equation. Subtract \frac{1}{4} from both sides.

$$\frac{7}{8}b=1-\frac{1}{4}$$

Subtract \frac{1}{4} from 1 to get \frac{3}{4}.

$$\frac{7}{8}b=\frac{3}{4}$$

Multiply both sides by \frac{8}{7}, the reciprocal of \frac{7}{8}.

$$b=\frac{3}{4}\times \frac{8}{7}$$

Multiply \frac{3}{4} and \frac{8}{7} to get \frac{6}{7}.

$$b=\frac{6}{7}$$

The system is now solved.

$$b=\frac{6}{7} c=2 d=2$$