1. ## Simultenous Equations

How to solve $x+\sqrt{y}=4$ and $y+\sqrt{x}=3$ Thanks in advance. x=? y=?

2. Originally Posted by JohnDoe
How to solve $x+\sqrt{y}=4$ and $y+\sqrt{x}=3$ Thanks in advance. x=? y=?
rearranging:

$\sqrt{y} = 4-x$ so $y = (4-x)^2 = 16-8x+x^2$

$\sqrt{x} = 3-y$ so $x = (3-y)^2 = 9-6y+y^2$

Sub in (4-x)^2 for y:

$x = (3-16+8x-x^2)^2$

solve for x and the put the value(s) of x into one of the original equations to find y

3. xsoy I did not get the meaning. Can you explain?

Thanks for reply but I can not figure out how to do it can anyone provide me the exact solution. Also I am not experienced with MathType so I had to edit the post a few times the correct one would be $x+\sqrt{y}=4$ and $y+\sqrt{x}=3$.

There must be a trick in solving this problem because it is an examination question and we are not supposed to use calculators while solving with the method you suggest I think it is impossible to get an accurate solution without a calculator.