Removing a radical from an addend in an equation?

Question

Given this simple equation:

$$ a + b\sqrt{5} = 7\sqrt{5} - 7 $$

I can really easily say that $a = -7$ and $b = 7$, because:

$ a + b\sqrt{5} = 7\sqrt{5} - 7 \\ b\sqrt{5} + a = 7\sqrt{5} - 7 \Rightarrow b = 7 \land a = -7 $

But, what if I want to show it with step-by-step passages?

I know I would just need to find out that $\,\,\,a + b = 0 \,\,\,$ and then I would say that $\,\,a = -b \,\,$ and substitute $\,a\,$ in the original equation:

$ a + b\sqrt{5} = 7\sqrt{5} - 7 = -b + b\sqrt{5} \\ b(\sqrt{5} - 1) = 7\sqrt{5} - 7 \\ b = \frac{7\sqrt{5} - 7}{\sqrt{5} - 1} \\ b = \frac{7\sqrt{5} - 7}{\sqrt{5} - 1} \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = 7 \\ a = -b = -7 $

But starting from:

$a + b\sqrt{5} = 7\sqrt{5} - 7$

I have an irrational radical which is multiplied by $b$, and if I put the equation into a system:

$$ \left\{ \begin{array}{c} a + b\sqrt{5} = 7\sqrt{5} - 7 \\ a = 7\sqrt{5} - 7 -b\sqrt{5}\\ \end{array} \right. $$ $$ \left\{ \begin{array}{c} 7\sqrt{5} - 7 -b\sqrt{5} + b\sqrt{5} = 7\sqrt{5} - 7 \\ 7\sqrt{5} - 7 = 7\sqrt{5} - 7 \\ \end{array} \right. $$

This is where it finishes. So the question is: is there a way to get rid of a radical which is a factor of only one addend, like in $a + b\sqrt{5}$ and obtain $a + b$? What are the passages?

Thanks for the attention.

Answer 1

First of all you are solving the equation for rational $a,b$ as otherwise this equation has an infinite set of solutions. Secondly you should know that rationals and irrationals are closed sets under multiplication by rational numbers. Therefore you can consider the irrational parts of the equation and the rational parts separately to solve for $a$ and $b$.

Answer 2

You can transform to $$ a+7=(7-b)\sqrt 5$$ and - unless $b=7$ - to $$ \sqrt 7=\frac{a+7}{7-b}.$$ If $a,b$ are rational this claims that $\sqrt 7$ is rational, which is absurd. Hence division by $7-b$ was not allowed after all, i.e., $b=7$ and then $a+7=0$, i.e., $a=-7$.