Solve for t t/4+3=2t-b

$\frac{t}{4} + 3 = 2 t - b$

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

Subtract $2 t$ from both sides of the equation.

$\frac{t}{4} + 3 - 2 t = - b$

To write $- 2 t$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{t}{4} - 2 t \cdot \frac{4}{4} + 3 = - b$

Combine $- 2 t$ and $\frac{4}{4}$ .

$\frac{t}{4} + \frac{- 2 t \cdot 4}{4} + 3 = - b$

Combine the numerators over the common denominator.

$\frac{t - 2 t \cdot 4}{4} + 3 = - b$

Simplify each term.

Simplify the numerator.

Factor $t$ out of $t - 2 t \cdot 4$ .

Raise $t$ to the power of $1$ .

$\frac{t^{1} - 2 t \cdot 4}{4} + 3 = - b$

Factor $t$ out of $t^{1}$ .

$\frac{t \cdot 1 - 2 t \cdot 4}{4} + 3 = - b$

Factor $t$ out of $- 2 t \cdot 4$ .

$\frac{t \cdot 1 + t (- 2 \cdot 4)}{4} + 3 = - b$

Factor $t$ out of $t \cdot 1 + t (- 2 \cdot 4)$ .

$\frac{t (1 - 2 \cdot 4)}{4} + 3 = - b$

Multiply $- 2$ by $4$ .

$\frac{t (1 - 8)}{4} + 3 = - b$

Subtract $8$ from $1$ .

$\frac{t \cdot - 7}{4} + 3 = - b$

Move $- 7$ to the left of $t$ .

$\frac{- 7 \cdot t}{4} + 3 = - b$

Move the negative in front of the fraction.

$- \frac{7 t}{4} + 3 = - b$

Subtract $3$ from both sides of the equation.

$- \frac{7 t}{4} = - b - 3$

Multiply both sides of the equation by $- \frac{4}{7}$ .

$- \frac{4}{7} (- \frac{7 t}{4}) = - \frac{4}{7} (- b - 3)$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{4}{7} (- \frac{7 t}{4})$ .

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{4}{7}$ into the numerator.

$\frac{- 4}{7} (- \frac{7 t}{4}) = - \frac{4}{7} (- b - 3)$

Move the leading negative in $- \frac{7 t}{4}$ into the numerator.

$\frac{- 4}{7} \cdot \frac{- 7 t}{4} = - \frac{4}{7} (- b - 3)$

Factor $4$ out of $- 4$ .

$\frac{4 (- 1)}{7} \cdot \frac{- 7 t}{4} = - \frac{4}{7} (- b - 3)$

Cancel the common factor.

$\frac{4 \cdot - 1}{7} \cdot \frac{- 7 t}{4} = - \frac{4}{7} (- b - 3)$

Rewrite the expression.

$\frac{- 1}{7} (- 7 t) = - \frac{4}{7} (- b - 3)$

Cancel the common factor of $7$ .

Factor $7$ out of $- 7 t$ .

$\frac{- 1}{7} (7 (- t)) = - \frac{4}{7} (- b - 3)$

Cancel the common factor.

$\frac{- 1}{7} (7 (- t)) = - \frac{4}{7} (- b - 3)$

Rewrite the expression.

$- - t = - \frac{4}{7} (- b - 3)$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 t = - \frac{4}{7} (- b - 3)$

Multiply $t$ by $1$ .

$t = - \frac{4}{7} (- b - 3)$

Simplify the right side.

Simplify $- \frac{4}{7} (- b - 3)$ .

Apply the distributive property.

$t = - \frac{4}{7} (- b) - \frac{4}{7} \cdot - 3$

Multiply $- \frac{4}{7} (- b)$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{4}{7}) b - \frac{4}{7} \cdot - 3$

Multiply $\frac{4}{7}$ by $1$ .

$t = \frac{4}{7} b - \frac{4}{7} \cdot - 3$

Combine $\frac{4}{7}$ and $b$ .

$t = \frac{4 b}{7} - \frac{4}{7} \cdot - 3$

Multiply $- \frac{4}{7} \cdot - 3$ .

Multiply $- 3$ by $- 1$ .

$t = \frac{4 b}{7} + 3 (\frac{4}{7})$

Combine $3$ and $\frac{4}{7}$ .

$t = \frac{4 b}{7} + \frac{3 \cdot 4}{7}$

Multiply $3$ by $4$ .

$t = \frac{4 b}{7} + \frac{12}{7}$

Solve for t t/4+3=2t-b

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