# GRE basic algebra problem, plugging in works, algebreic method fails me

Simple GRE practice problem, but for some reason my algebraic approach is failing me, can someone point out my error?

Given: $$\theta x = x^{-3}(2x)(\frac{x}{2})(2)$$

Question: Which is greater: $\theta 8$ or $\theta 4$

I approached it by solving for theta:

$$\theta x = x^{-3}(2x)(x)$$ $$\theta x = \frac{2x^2}{x^3}$$ $$\theta x = \frac{2}{x}$$ $$\theta = \frac{2}{x^2}$$

Then plug in for $\theta 8$ and $\theta 4$: $$\theta * 8 = \frac{2}{x^2} (8) = \frac{16}{x^2}$$ $$\theta * 4 = \frac{2}{x^2} (4) = \frac{8}{x^2}$$

For any value of $x$, other than $0$, $\theta 8$ is larger. But clearly amiss here, because the opposite is true, if I plug in $\theta 8$ directly I get:

$$\theta 8 = 8^{-3}(2*8)(\frac{8}{2})(8) = \frac{1}{4}$$ $$\theta 4 = 4^{-3}(2*4)(\frac{4}{2})(4) = \frac{1}{2}$$

Now it's clear that $\theta 4$ is larger.

Ooff! for the life of me I don't see why my algebraic approach failed. How'd I get myself into this quandary? And more importantly, how do I get out using algebra?

• $\theta(8)$ can't end up with an $x$ in it. $x=8$. – Thomas Andrews Nov 4 '14 at 22:47
• I didn't mean to write theta(8) as a function, I meant the notation as theta*8 (a coincidentally ambiguous notation at an inopportune moment) – David Parks Nov 4 '14 at 23:30
• But none of the above makes sense as $\theta * x$, it only makes sens as $\theta(x)$, a function. – Thomas Andrews Nov 4 '14 at 23:31
• The practice question (out of Princeton Review) gave the equation in the form exactly as shown here, and it was a simple which is greater question with the same form shown $\theta 8$ or $\theta 4$. I was simply plugging in the equivalent form of theta (which I got from simplifying the given equation) into the statement $\theta 8$. – David Parks Nov 4 '14 at 23:37
• But then $\theta$ would be a constant, which makes no sense, because no constant fits for all $x$. If you don't agree it means a function, why did you select Nick's answer below as "correct," when it says that $\theta$ is a function? – Thomas Andrews Nov 4 '14 at 23:39

This is a functions question, where $\theta x$ represents we are taking $x$ as an input to the function theta. Think of it as $f(x)=x^{-3}(2x)(\dfrac{x}{2})(2)$ instead. You cannot "solve" for theta because it is just notation.

• Thanks for the answer. I see what your saying, as a function this all makes sense. I can't say I quite internalize the rational here though. I still look at it with some bewilderment. (1) I see an equation. (2) I manipulated that equation without (knowing) violating any laws of basic mathematics. (3) I now have an equivalency for theta. (4) Shouldn't I be able to replace theta with its equivalent form safely? – David Parks Nov 4 '14 at 23:34
• I actually have an interest in math well above this level, so mucking this up really scares the bajeebers out of me, and I want to make sure I get this simple foundation straight in my head. :) – David Parks Nov 4 '14 at 23:35
• Given how the question is written, you should think of $\theta$ as a variable as you were doing. That's why both NickC and I are guessing it's a typo and should be expressed as $\theta(x)$ rather than $\theta x$. So either you made a mistake in transcribing the question on this site, or you transcribed it correctly and are right to be confused - it was a typo wherever you saw it! – Shane Nov 5 '14 at 1:36

Maybe it should be $$\theta(x)=x^{-3}(2x)(x/2)2 = 2x^{-1} = \frac{2}{x}$$

Then obviously $\theta(4)$ > $\theta(8)$.

GRE's are not tough, but that seems too easy. Am I missing something?

• Can I ask what a GRE is just for my own curiosity? :) – snulty Nov 5 '14 at 0:19
• Graduate record examination. The test you take to get into grad school. – Shane Nov 5 '14 at 0:51