1. ## System of equations

Hi,

My question is how do you solve analitically:

sqrt(x)+y=7 and sqrt(y)+x=11 without solving a 4th degree polynomial equation?

Thanks

2. Originally Posted by kolg
Hi,

My question is how do you solve:

sqrt(x)+y=7 and sqrt(y)+x=11 without solving a 4th degree polynomial equation?

Thanks
graphing ...

3. Dear kolg,

$\sqrt{x}+y=11$------------(1)

$\sqrt{y}+x=7$-------------(2)

From (1); y<11 and $x\geq{0}$

From (2); x<7 and $y\geq{0}$

Therefore, $0\leq{y}<11$ and $0\leq{x}<7$

Also from (1) and (2) it could be seen that, $\sqrt{x}\in{N}$ and $\sqrt{y}\in{N}$

Therefore x=4 is the only solution for x.

When, $x=4\Rightarrow{y=9}$

Therefore x=4, y=9 is the only solution for this system of equations.

4. Hi

Thanks for the reply but how do you know that srqrt(x) and sqrt(y) are integers? Can't they be racionals or irracionals?

Thanks

5. Dear Klog,

If $\sqrt{x}$ is rational but not an integer,

$\sqrt{x}=\frac{p}{q}$ where $q\neq{1}$

$x=\frac{p^2}{q^2}$

Therefore, $\sqrt{y}=7-\frac{p^2}{q^2}=(\sqrt{7}-\frac{p}{q})(\sqrt{7}-\frac{p}{q})$

Hence $\sqrt{y}$ is irrational.

Then $x+\sqrt{y}$ is irrational. (Contradiction)

Therefore $\sqrt{x}$ is irrational or an integer.(according to our assumption.)

But if $\sqrt{x}$ is irrational $y+\sqrt{x}$ is irrational. (Contradiction)

So $\sqrt{x}$ is an integer.

Same proof goes to $\sqrt{y}$ as well.

Hope this helps.