Answer:
[tex]\huge\boxed{k=21}[/tex]
Step-by-step explanation:
[tex]\dfrac{9}{2k-3}=\dfrac{4}{k+1}[/tex]
First step:
Find domain.
We know: the denominator must be different than 0.
Therefore we have:
[tex]2k-3\neq0\ \wedge\ k+1[/tex]
[tex]2k-3\neq0\qquad\text{add 3 to both sides}\\2k\neq3\qquad\text{divide both sides by 2}\\\boxed{k\neq1.5}\\\\k+1\neq0\qquad\text{subtract 1 from both sides}\\\boxed{k\neq-1}\\\\\text{Domain:}\ x\in\mathbb{R}\backslash\{-1;\ 1.5\}[/tex]
Second step:
Solve for k.
[tex]\dfrac{9}{2k-3}=\dfrac{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}[/tex]
