Is there a method where I'm able to solve $7-r-2k=0$ without trial and error?

Question

unklerhaukus · Answer

solve for r(k)
or k(r)?

anonymous · Answer

solve for both $r$ and $k$

anonymous · Answer

like finding both variable.

unklerhaukus · Answer

$$ 7−r−2k=0 $$
$$ r=7−2k $$$$k=\frac{r-7}2$$

anonymous · Answer

Omniscience, this equation doesn't have a single answer. It has infinitely many answers, which is why we can't simply say, "Oh, r equals this, and k equals this." So instead, what we can do is exactly what my boy Rhaukus suggested. We can find an equation r(k) or k(r) which means then that we can pick any k, plug it into the equation to get r, and there would be ONE of the many solutions.

unklerhaukus · Answer

$$*\qquad k=\frac{7-r}2$$

anonymous · Answer

okay, then i can just the first variable and find the next?

anonymous · Answer

find the first*

anonymous · Answer

So, Rhaukus solved for r and got:
r = 7-2k

To find an individual solution, I'll pick any k that I want. I want to pick k=5, just because. I plug k into my equation:
r = 7-2(5)
r= 7-10
r=-3

So ONE solution of the equation is
k=5, r=-3

unklerhaukus · Answer

the value of the variables depend on each other

anonymous · Answer

But its from this..
$$\left(2x+3\right)^{4}  \left(x-\frac{2}{x}\right)  ^{2}   $$
$$\left(\begin{matrix}4 \ r\end{matrix}\right)*\left(\begin{matrix}3 \ k\end{matrix}\right)2^{4-r}3^{r}(-2)^{k}x^{7-r-2k}$$
where im finding the variable of $r$ and $k$

anonymous · Answer

ahaha oh my god.

anonymous · Answer

which means im only able to use trial and error? because you would never know which number to plug in..

unklerhaukus · Answer

substitute   $r=7-2k$    into this

unklerhaukus · Answer

$$\left(\begin{matrix}4 \ 7−2k \end{matrix}\right)*\left(\begin{matrix}3 \ k\end{matrix}\right)2^{4-7−2k }3^{7−2k }(-2)^{k}x^{7-(7-2k)-2k}$$

unklerhaukus · Answer

well the coefficient of x can be simplified

anonymous · Answer

is that method allowed?

unklerhaukus · Answer

i dont know the context,
im not even sure what is in those big brackets...

anonymous · Answer

well, im finding the constant term for the expansion..

unklerhaukus · Answer

does it depend on x?

anonymous · Answer

I am just wondering; if there is a method where I don't need to use trial and error because it's long. Thanks anyway!

anonymous · Answer

The constant term does not have $x$ in it; the independent term