Solve for b 8(3b+1)=4(b+3)-9

$8 (3 b + 1) = 4 (b + 3) - 9$

Step 1

Simplify $8 (3 b + 1)$ .

Rewrite.

$0 + 0 + 8 (3 b + 1) = 4 (b + 3) - 9$

Simplify by adding zeros.

$8 (3 b + 1) = 4 (b + 3) - 9$

Apply the distributive property.

$8 (3 b) + 8 \cdot 1 = 4 (b + 3) - 9$

Multiply.

Multiply $3$ by $8$ .

$24 b + 8 \cdot 1 = 4 (b + 3) - 9$

Multiply $8$ by $1$ .

$24 b + 8 = 4 (b + 3) - 9$

Step 2

Simplify $4 (b + 3) - 9$ .

Simplify each term.

Apply the distributive property.

$24 b + 8 = 4 b + 4 \cdot 3 - 9$

Multiply $4$ by $3$ .

$24 b + 8 = 4 b + 12 - 9$

Subtract $9$ from $12$ .

$24 b + 8 = 4 b + 3$

Step 3

Move all terms containing $b$ to the left side of the equation.

Subtract $4 b$ from both sides of the equation.

$24 b + 8 - 4 b = 3$

Subtract $4 b$ from $24 b$ .

$20 b + 8 = 3$

Step 4

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

Subtract $8$ from both sides of the equation.

$20 b = 3 - 8$

Subtract $8$ from $3$ .

$20 b = - 5$

Step 5

Divide each term in $20 b = - 5$ by $20$ and simplify.

Divide each term in $20 b = - 5$ by $20$ .

$\frac{20 b}{20} = \frac{- 5}{20}$

Simplify the left side.

Cancel the common factor of $20$ .

Cancel the common factor.

$\frac{20 b}{20} = \frac{- 5}{20}$

Divide $b$ by $1$ .

$b = \frac{- 5}{20}$

Simplify the right side.

Cancel the common factor of $- 5$ and $20$ .

Factor $5$ out of $- 5$ .

$b = \frac{5 (- 1)}{20}$

Cancel the common factors.

Factor $5$ out of $20$ .

$b = \frac{5 \cdot - 1}{5 \cdot 4}$

Cancel the common factor.

$b = \frac{5 \cdot - 1}{5 \cdot 4}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$b = - 0.25$

Solve for b 8(3b+1)=4(b+3)-9

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