Solving complex algebraic equation with radicals

Question

How do I solve this math question? If $x$ and $y$ are rational numbers and $(3 + 4\sqrt{3})(x + y\sqrt{3}) = 26$, find the sum of $x$ and $y$.

I tried solving for $x$ and $y$ individually to add them up, but haven't been able to obtain a rational number for either variable as I can't seem to get rid of the $\sqrt3$.

Thank you for your help.

Answer 1

The key here is that $x$ and $y$ must be rational.

Expanding the brackets, we get:

$$ \begin{align} 26 &=(3 + 4\sqrt{3})(x + y\sqrt{3}) \\ &=3x + 4x\sqrt{3}+3y\sqrt{3}+12y \\ &= 3(x+4y) + \sqrt{3}(4x+3y) \tag{1} \end{align} $$

Now, the $26$ on the left hand side is obviously rational. Since $x$ and $y$ must also be rational, the only way to make the right hand side rational is to have:

$$4x+3y=0 \tag{2}$$

In this case, $(1)$ becomes:

$$x+4y=\frac{26}{3} \tag{3}$$

Now you could just solve to find $x$ and $y$ now. But a fancier way is to multiply equation $(2)$ by $3$ to get:

$$12x+9y=0 \tag{4}$$

Then add equations $(3)$ and $(4)$ to get:

$$13x+13y=\frac{26}{3} \tag{5}$$

... and I'll leave the dividing by $13$ to you!

Answer 2

What you need to do is expand the product. You will have some terms like $3x$ which is rational, and terms like $3y\sqrt{3}$ which is not. Since the term on the right is rational, it means that the sum of irrational terms is $0$, and the sum of rational terms is $26$

Answer 3

Alt. hint: $\;x + y\sqrt{3} = \dfrac{26}{3 + 4\sqrt{3}} \color{red}{\cdot \dfrac{4 \sqrt{3}-3}{4\sqrt{3}-3}}=\dfrac{26(4 \sqrt{3}-3)}{39}=\dfrac{8 \sqrt{3}-6}{3}=-2+\dfrac{8}{3}\sqrt{3}\,$.