Solve A=7/8(B+C)

$$A = \frac { 7 } { 8 } ( B + C )$$

Solve for B

$$B=\frac{8A}{7}-C$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{7}{8} by B+C.

$$A=\frac{7}{8}B+\frac{7}{8}C$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{7}{8}B+\frac{7}{8}C=A$$

Subtract \frac{7}{8}C from both sides.

$$\frac{7}{8}B=A-\frac{7}{8}C$$

The equation is in standard form.

$$\frac{7}{8}B=-\frac{7C}{8}+A$$

Divide both sides of the equation by \frac{7}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{7}{8}B}{\frac{7}{8}}=\frac{-\frac{7C}{8}+A}{\frac{7}{8}}$$

Dividing by \frac{7}{8} undoes the multiplication by \frac{7}{8}.

$$B=\frac{-\frac{7C}{8}+A}{\frac{7}{8}}$$

Divide A-\frac{7C}{8} by \frac{7}{8} by multiplying A-\frac{7C}{8} by the reciprocal of \frac{7}{8}.

$$B=\frac{8A}{7}-C$$

Solve for A

$$A=\frac{7\left(B+C\right)}{8}$$

Solution Steps

Use the distributive property to multiply \frac{7}{8} by B+C.

$$A=\frac{7}{8}B+\frac{7}{8}C$$