I'm trying to come up with a complicated equation that equals 53.
5x - 2x + x * 7x + 23 * x / 78652
x = 53
That one is pretty lackluster though.
I'm looking forward to seeing what the math wizards here can come up with.
I'm trying to come up with a complicated equation that equals 53.
5x - 2x + x * 7x + 23 * x / 78652
x = 53
That one is pretty lackluster though.
I'm looking forward to seeing what the math wizards here can come up with.
I mean one can make this as arbitrarily complicated as you want.
For example, even the coefficents in your above equation can be complicated amazingly.
$\displaystyle 2=\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n=\s um_{n=0}^{\infty}e^{n\ln\left(\frac{1}{2}\right)}= \sum_{n=0}^{\infty}\sum_{\ell=0}^{\infty}\frac{\le ft(n\ln\left(\frac{1}{2}\right)\right)^{\ell}}{\el l!}\cdots$ ad infinitum
Unless you clearly state exactly what mathematical background can be drawn upon, this question is just going to cause confusion and waste time. Note: Posting this question in Pre-algebra and Algebra implies a background severly limit what you might otherwise have in mind.
I should've written it differently.
I'm wanting to say: Y + X = 53
5x - 2x + x * 7x + 23 * x / 78652
Something like that.
I guess shouldn't have left the question so open. I'm looking for something that is algebra and can be solved in say 10 minutes by someone with a good grasp of algebra.
Hello, Jerick!
I'm trying to come up with a complicated equation that equals 53.
So we given with: .$\displaystyle x \:=\:53$
Subtract 3 from both sides: .$\displaystyle x - 3 \:=\:50$
Double both sides: .$\displaystyle 2(x-3) \:=\:100$
Subtract 9 from both sides: .$\displaystyle 2(x-3) - 9 \:=\:91$
Divide by 7: .$\displaystyle \boxed{\frac{2(x-3) = 9}{7} \:=\:13}$
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If $\displaystyle x = 5$ can be one of the roots of the equation . . .
Start with: .$\displaystyle x \:=\:53$
Subtract 3: .$\displaystyle x - 3 \:=\:50$
Divide by 5: .$\displaystyle \frac{x-3}{3} \:=\:10$
Cube both sides: .$\displaystyle \left(\frac{x-3}{5}\right)^3 \:=\:10^3 \quad\Rightarrow\quad \frac{(x-3)^3}{125} \:=\:1000$
. . $\displaystyle x^3 - 9x^2 + 27x - 27 \:=\:125,\!000 \quad\Rightarrow\quad\boxed{ x^3 - 9x^2 + 27x - 125,\!027 \:=\:0} $