Solve for k 9/(2k-3)=4/(k+1)

$\frac{9}{2 k - 3} = \frac{4}{k + 1}$

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.

$9 (k + 1) = (2 k - 3) \cdot 4$

Step 2

Solve the equation for $k$ .

Simplify $9 (k + 1)$ .

Rewrite.

$0 + 0 + 9 (k + 1) = (2 k - 3) \cdot 4$

Simplify by adding zeros.

$9 (k + 1) = (2 k - 3) \cdot 4$

Apply the distributive property.

$9 k + 9 \cdot 1 = (2 k - 3) \cdot 4$

Multiply $9$ by $1$ .

$9 k + 9 = (2 k - 3) \cdot 4$

Simplify $(2 k - 3) \cdot 4$ .

Apply the distributive property.

$9 k + 9 = 2 k \cdot 4 - 3 \cdot 4$

Multiply.

Multiply $4$ by $2$ .

$9 k + 9 = 8 k - 3 \cdot 4$

Multiply $- 3$ by $4$ .

$9 k + 9 = 8 k - 12$

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

Subtract $8 k$ from both sides of the equation.

$9 k + 9 - 8 k = - 12$

Subtract $8 k$ from $9 k$ .

$k + 9 = - 12$

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

Subtract $9$ from both sides of the equation.

$k = - 12 - 9$

Subtract $9$ from $- 12$ .

$k = - 21$

Solve for k 9/(2k-3)=4/(k+1)

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