Solve for t k=(2(t+3))/(t-3)

$k = \frac{2 (t + 3)}{t - 3}$

Multiply the equation by $t - 3$ .

$(t - 3) (k) = (\frac{2 (t + 3)}{t - 3}) (t - 3)$

Simplify the left side.

Apply the distributive property.

$t k - 3 k = (\frac{2 (t + 3)}{t - 3}) (t - 3)$

Simplify the right side.

Simplify $(\frac{2 (t + 3)}{t - 3}) (t - 3)$ .

Cancel the common factor of $t - 3$ .

Cancel the common factor.

$t k - 3 k = \frac{2 (t + 3)}{t - 3} (t - 3)$

Rewrite the expression.

$t k - 3 k = 2 (t + 3)$

Apply the distributive property.

$t k - 3 k = 2 t + 2 \cdot 3$

Multiply $2$ by $3$ .

$t k - 3 k = 2 t + 6$

Solve for $t$ .

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

$t k - 3 k - 2 t = 6$

Add $3 k$ to both sides of the equation.

$t k - 2 t = 6 + 3 k$

Factor $t$ out of $t k - 2 t$ .

Factor $t$ out of $t k$ .

$t (k) - 2 t = 6 + 3 k$

Factor $t$ out of $- 2 t$ .

$t (k) + t \cdot - 2 = 6 + 3 k$

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

$t (k - 2) = 6 + 3 k$

Divide each term in $t (k - 2) = 6 + 3 k$ by $k - 2$ and simplify.

Divide each term in $t (k - 2) = 6 + 3 k$ by $k - 2$ .

$\frac{t (k - 2)}{k - 2} = \frac{6}{k - 2} + \frac{3 k}{k - 2}$

Simplify the left side.

Cancel the common factor of $k - 2$ .

Cancel the common factor.

$\frac{t (k - 2)}{k - 2} = \frac{6}{k - 2} + \frac{3 k}{k - 2}$

Divide $t$ by $1$ .

$t = \frac{6}{k - 2} + \frac{3 k}{k - 2}$

Simplify the right side.

Combine the numerators over the common denominator.

$t = \frac{6 + 3 k}{k - 2}$

Factor $3$ out of $6 + 3 k$ .

Factor $3$ out of $6$ .

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

Factor $3$ out of $3 \cdot 2 + 3 k$ .

$t = \frac{3 (2 + k)}{k - 2}$

Solve for t k=(2(t+3))/(t-3)

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