Question

Solve the system.

√3 x-√2 y=2√3
2√2 x√3 y=√2

Answer 1

Sorry I just realized I left a + out 2√2 x+√3 y=√2

Answer 2

What is preventing you from solving this system?

Can you solve this one?

3x - 2y=2(3)
2(2)x + 3y=2

Answer 3

These symbols: [1], [2], [3], etc. denote the equation number for better reference.
\[\sqrt{3} x-\sqrt{2} y=2\sqrt{3}[1]\]
\[2\sqrt{2} x +\sqrt{3} y=\sqrt{2}[2]\]
\[[1]\times \frac{\sqrt{3}}{\sqrt{2}}\]
\[\large \frac{\sqrt{3}}{\sqrt{2}}[\sqrt{3} x-\sqrt{2} y]=\frac{\sqrt{3}}{\sqrt{2}}[2\sqrt{3}]\]
\[\large \frac{3}{\sqrt{2}}x-\sqrt{3} y=\frac{2(3)}{\sqrt{2}}\]
\[\large \frac{3}{\sqrt{2}}x-\sqrt{3} y=\frac{6}{\sqrt{2}}[3]\]
\[[2]+[3]\]
\[\large [2\sqrt{2} x +\sqrt{3} y]+[\frac{3}{\sqrt{2}}x-\sqrt{3} y]=[\sqrt{2}]+[\frac{6}{\sqrt{2}}]\]
\[\large 2\sqrt{2}x+\frac{3}{\sqrt{2}}x=\frac{2+6}{\sqrt{2}}\]
\[\large x(2\sqrt{2}+\frac{3}{\sqrt{2}})=\frac{8}{\sqrt{2}}\]
\[\large x=\frac{8}{\sqrt{2}} \times \frac{1}{2\sqrt{2}+\frac{3}{\sqrt{2}}}\]
\[\large x=\frac{8}{\sqrt{2}[2\sqrt{2}+\frac{3}{\sqrt{2}}]}\]
\[\large x=\frac{8}{2(2)+3}\]
\[\x=\frac{8}{7}\]

Answer 4

Find the y-value by substituting the x-value into either equation [1] or [2] or [3]

Answer 5

so there is no nicer way to solve this but using a bunch of √√√?

Answer 6

I don't see any difficulty when there's square root signs. Square root signs aren't scary if you just apply your basic addition multiplication with them and everything will turn out fine. But I see no other method or any other means to solve this except maybe matrix or substitution, which would even be more tedious.

Answer 7

If look closely at my method, you would be thinking to yourself, hey, that wasn't so bad. If you follow it through you will have that sense that it wasn't that bad. But if you're not that good at adding or dividing fractions, then maybe you will struggle a bit. I tried to make it very clear by denoting symbols ie. [1] means equation 1, etc as well as giving you a step by step solution.

Answer 8

If you look*

Answer 9

If you have any problems with that, then you can tag me in here and I will see what I can do. But other people in here may think of a better solution but this is how I usually do it if there is no variable that's monic.

Answer 10

Thanks Azteck. I'm copying your solution and will go through it step by step and see if i have any issues or questions.

Answer 11

?? No, I was not suggesting that the solution would be the same without the radicals. I was only hoping to suggest that the coefficients don't matter. Numbers are numbers. Why is it more difficult or confusing with or without radicals?

Answer 12

@tkhunny
When I start working with radicals and fractions, its all just looks way over complicated. it is solvable, it looks ugly, i was just hoping there is a better way to do this, maybe get rid of the radicals or something like that.

Answer 13

Doesn't look ugly when radicals cancel out and you're left with whole numbers.