How to solve $\displaystyle x+\sqrt{y}=4$ and $\displaystyle y+\sqrt{x}=3$ Thanks in advance. x=? y=?
rearranging:
$\displaystyle \sqrt{y} = 4-x$ so $\displaystyle y = (4-x)^2 = 16-8x+x^2$
$\displaystyle \sqrt{x} = 3-y$ so $\displaystyle x = (3-y)^2 = 9-6y+y^2$
Sub in (4-x)^2 for y:
$\displaystyle 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
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 $\displaystyle x+\sqrt{y}=4$ and $\displaystyle y+\sqrt{x}=3$.
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.