Question

Solve for k. -21 -3 21

Answer 1

`\huge\boxed{k=21}`

Explanation:

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

First step:

Find domain.

We know: the denominator must be different than 0.

Therefore we have:

`2k-3\\eq0\ \wedge\ k+1`

`2k-3\\eq0\qquad\text{add 3 to both sides}\\2k\\eq3\qquad\text{divide both sides by 2}\\\boxed{k\\eq1.5}\\\\k+1\\eq0\qquad\text{subtract 1 from both sides}\\\boxed{k\\eq-1}\\\\\text{Domain:}\ x\in\mathbb{R}\backslash\{-1;\ 1.5\}`

Second step:

Solve for k.

`(9)/(2k-3)=(4)/(k+1)\qquad\text{cross multiply}\\\\9(k+1)=4(2k-3)\qquad\text{use the distributive property}:\ a(b+c)=ab+ac\\\\(9)(k)+(9)(1)=(4)(2k)-(4)(3)\\\\9k+9=8k-12\qquad\text{subtract 9 from both sides}\\\\9k=8k-21\qquad\text{subtract}\ 8k\ \text{from both sides}\\\\\boxed{k=21}\in\text{Domain}`

Answer 2

k = -21

Explanation:

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

Using cross products

9 * (k+1) = 4(2k-3)

Distribute

9k+9 = 8k -12

Subtract 8k from each side

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

k+9 = -12

Subtract 9 from each side

k+9-9 = -12-9

k = -21