Solve for k 9/(k+3)=3/5

$\frac{9}{k + 3} = \frac{3}{5}$

Step 1

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$9 \cdot 5 = (k + 3) \cdot 3$

Step 2

Solve the equation for $k$ .

Rewrite the equation as $(k + 3) \cdot 3 = 9 \cdot 5$ .

$(k + 3) \cdot 3 = 9 \cdot 5$

Multiply $9$ by $5$ .

$(k + 3) \cdot 3 = 45$

Divide each term in $((k + 3) \cdot 3) = 45$ by $3$ and simplify.

Divide each term in $((k + 3) \cdot 3) = 45$ by $3$ .

$\frac{((k + 3) \cdot 3)}{3} = \frac{45}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{(k + 3) \cdot 3}{3} = \frac{45}{3}$

Divide $k + 3$ by $1$ .

$k + 3 = \frac{45}{3}$

Simplify the right side.

Divide $45$ by $3$ .

$k + 3 = 15$

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

Subtract $3$ from both sides of the equation.

$k = 15 - 3$

Subtract $3$ from $15$ .

$k = 12$

Solve for k 9/(k+3)=3/5

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