Suppose $a$ and $b$ are integers in the ratio $6:7$, what is the ratio of $(6a^{2} + 7b^{2} + 6a + 7b ) : (a^{2}+b^2)$?

Question

Initially I thought I can assume $a=6$ and $b=7$ and calculate the ratio of the given expression which turned out to be $\frac{644}{85}$ and this was not the correct answer provided for this question.

Then when I turned to the solution for this problem, this statement was written as part of the solution

The degree of all the terms in the numerator is not equal and hence, the resultant ratio can't be computed.

I couldn't understand this statement. What does it mean? What's the theory behind this? I am just a little confused here and I am sorry if this question looks foolish but a little guidance will be helpful.

Answer 1

I get a ratio of $\frac{7}{b}+\frac{559}{85}$ when calculating $a=\frac{6 b}{7}$ and $\frac{6 a^2+6 a+7 b^2+7 b}{a^2+b^2}$.

In turn, when setting $b=\frac{7 a}{6}$, I obtain a ratio of $\frac{6}{a}+\frac{559}{85}$.

The fact that squared variables $a^2$, $b^2$ are involved leads to a non-constant ratio (where at least one variable $a$ or $b$ is involved). If you use only linear terms in the denominator and numerator, we obtain a constant as ratio.

And when I use cubic terms, for example $\frac{6 a^3+6 a+7 b^3+7 b}{a^3+b^3}$, I get a ratio depending from a squared $a$ or $b$, namely in this case $\frac{1}{559} \left(\frac{4165}{b^2}+3697\right)$.

We can generalize this question by setting $a=i$ and $b=j$ and use an arbitary exponent $n$ instead of $2$ which leads to a ratio:

$\frac{j \left(b^n+b\right)+\frac{b i^2}{j}+i \left(\frac{b i}{j}\right)^n}{b^n+\left(\frac{b i}{j}\right)^n}=\frac{ia^n+jb^n+ia+jb}{a^n+b^n}$ where $\frac{a}{b}=\frac{i}{j}$

Answer 2

The integers $6$ and $7$ are in a ratio $6:7$, and plugging them in yields a ratio $644:85$. The integers $-6$ and $-7$ are also in a ratio $6:7$, and plugging them yields a ratio $474:85$. These two ratios are clearly not the same, so the ratio is not determined by the ratio of $a$ and $b$. It is a trick question.

Answer 3

In $(6a^{2} + 7b^{2} + 6a + 7b ) : (a^{2}+b^2)$

we have the first two terms of same second degree homogenity so its ratio can be evaluated but the inhomogeneous second part of degree $1/L$ in linear units..

$$ \dfrac{ 6a + 7b}{a^{2}+b^2}$$

is different.. i.e. if we let $a:b= \dfrac{a}{b}=t\;$ it does not cancel, and so it cannot be evaluated in the same or in any other way.