Solve for b 63=7(1+4b)

$63 = 7 (1 + 4 b)$

Rewrite the equation as $7 (1 + 4 b) = 63$ .

$7 (1 + 4 b) = 63$

Divide each term in $7 (1 + 4 b) = 63$ by $7$ and simplify.

Divide each term in $7 (1 + 4 b) = 63$ by $7$ .

$\frac{7 (1 + 4 b)}{7} = \frac{63}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 (1 + 4 b)}{7} = \frac{63}{7}$

Divide $1 + 4 b$ by $1$ .

$1 + 4 b = \frac{63}{7}$

Simplify the right side.

Divide $63$ by $7$ .

$1 + 4 b = 9$

Move all terms not containing $b$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$4 b = 9 - 1$

Subtract $1$ from $9$ .

$4 b = 8$

Divide each term in $4 b = 8$ by $4$ and simplify.

Divide each term in $4 b = 8$ by $4$ .

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $b$ by $1$ .

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

Simplify the right side.

Divide $8$ by $4$ .

$b = 2$