# Where would this be on a range? ($0$ to $100\%$) Algebra [closed]

Ok, we have two expressions:

$30x + 1500$
$19x + 9000$

The first thing to note is, the first expression is $100\%$ and the second expression is $0\%$.

The second thing is to note in all of this is, $x$ is infinite in these expressions.

Ok, my main question is where would $23x + 15590$ go in the range as a percent? and how do you prove this? I don't know how but I'll accept any answer that makes sense. Bearing in mind $x$ is infinite, how?

EDIT: Hello guys, I apologize if that didn't make any sense. I meant in terms of this: link

## closed as unclear what you're asking by Roland, Martin R, almagest, Andreas Cap, Jendrik StelznerMay 19 '16 at 14:28

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• I have no idea what this means. 100% of what? If $x$ is infinite, then both expressions are infinite. – almagest May 19 '16 at 12:56
• If $x$ is infinite then multiplying it by a real or integer number or adding an integer or real number to it doesn't make much sense. And talking about their ratios (percentage is a form of expressing the ratio value) does not make sense at all, as $\frac \infty\infty$ is not defined in arithmetics. – CiaPan May 19 '16 at 12:58
• What does it mean to say that one expression is "0%" and another is "100%"? Are you looking for a linear progression from one to the other? – user247327 May 19 '16 at 12:58
• Moreover, these three lines aren't even parallel, so each intersects the other two. It's not clear what it means to propose some sort of "scale" to relate the third line to the other two. – colormegone May 19 '16 at 12:59
• If we ignore everything but the first coefficient for everything, we get $36.\overline{36}\%$ – Akiva Weinberger May 19 '16 at 13:16

I guess you mean three values linearly dependent on $x$, say $f(x), g(x), h(x),$ the third one falling between the former two for $x$ big enough; and you are asking about an asymptotic value of the proportion $$\frac{h-f}{g-f}$$ as $x$ grows to infinity.

That is \begin{align} \lim_{x\to\infty}\frac{(23x+15590)-(19x+9000)}{(30x+1500)-(19x+9000)} & = \lim_{x\to\infty}\frac{4x+6590}{11x-7500} \\ & = \lim_{x\to\infty}\frac{4+6590/x}{11-7500/x} \\ & = \frac{4+0}{11-0} \\ & = \frac 4{11} \\ & \approx 0,363636 \\ & \approx 36.4\% \end{align}

For example, here are some values of the fraction defined above:
for $x=10^3\ \,$ it is $\approx 3.025714286$,
for $x=10^4\ \,$ it is $\approx 0.454536586$,
for $x=10^8\ \,$ it is $\approx 0.363644834$ and
for $x=10^{10}$ it is $\approx 0.363636448$.

But please note that the limit value as $x$ grows to infinity is not the same as a 'proportion of infinite values'! The former exists and can be calculated, the latter simply does not exist, is undefined.

• @EthanBolker Thank you :) – CiaPan May 20 '16 at 6:16

Your objects are algebraic expressions. Among the possible interpretations are linear functions and straight lines in the plane. Interpreting them as linear functions would look like this: $$f(x) = 30x + 1500 \\ g(x) = 19x + 9000 \\ h(x) = 23x + 15590$$

The only thing that makes remotely sense to me is that you have some interpolation problem. $$\phi(1) = f \\ \phi(0) = g \\ \phi(\lambda) = h$$ where $\phi$ is a map that maps a parameter $\lambda$ to some linear function. E.g. $$\phi(\lambda) = \lambda f + (1-\lambda) g$$ would be such an interpolation. Alas your $h$ is probably not within the reachable functions of the above $\phi$.

Let us check it: $$\phi(\lambda) = \lambda(30x + 1500) + (1-\lambda)(19x +9000) \\ = (11 \lambda + 19) x + (9000 - 7500 \lambda)$$ comparison with $h$ gives the conditions $$11 \lambda + 19 = 23 \\ 9000 - 7500 \lambda = 15590$$ or $$\lambda = 4/11 > 0 \\ \lambda = -6590/7500 < 0$$ so there is no $\lambda \in \mathbb{R}$ which would result in $\phi(\lambda) = h$.