We have the fractions $\frac{n+3}{n-2}$ and $\frac{2n-3}{2n-8}$. Prove that these fractions are not equivalent (equal with each other)
My ideas and what I tried:
- I tried to multiply the terms $(n+3) \times (2n-8)$ and $(n-3) \times (2n-3)$ ,and said that they are equal and try to gave them another form and to show that the premise that they are equal is false (I don't know if this works and respect the math rules),
- another solving try was to equal $(n+3)$ with $(n-2)$, etc.)
- Another try was to transform them in a inequality equation.
With a single variable like $n$, I'm assuming you're trying to prove whether or not the equation is true for all real values of $n$. Therefore, if this were the case, then the desired solution is $n \in \mathbb{R}$.
To solve for $n$:
$\frac{n+3}{n-2} = \frac{n-2}{2n-8} \\ (n+3)(2n-8)=(n-2)(2n-3) \\ 2n^2-2n-24=2n^2-7n+6 \\ 5n = 30 \\ n = 6$
We can conclude that $n \notin \mathbb{R}$ and so the equation is not true for all values of $n$.