algebra: √x+y=11, X+√y=7; Find the value of X & Y

lalansingh said:
√x+y=11
X+√y=7
Find the value of X & Y
Click to expand...
As gently noted by the previous poster, "X" and "x" are two different variables, as are "Y" and "y". As currently posted, the exercise is unsolveable.

However, assuming that the different cases are typoes, the exercise appears to be to solve the following system of equations:

. . . . .\(\displaystyle \sqrt{x}\, +\, y\, =\, 11\)

. . . . .\(\displaystyle x\, +\, \sqrt{y}\, =\, 7\)

Since you've been assigned this exercise, obviously you've studied how to solve systems of equations (though possibly only linear ones, previously) and how to work with radical equations. So, based on that knowledge, what have you tried? How far have you gotten? Where are you stuck? For instance, you started by isolating the radicals in each equation, and squaring. Where did this lead?

Please be complete. Thank you! ;)
 
Given \(\displaystyle \sqrt{x}+ y= 11\) and \(\displaystyle x+ \sqrt{y}= 7\), the first thing I would do is let u= √x and v= √y so that the equations become u+ v^2= 11 and u^2+ v= 7. From the second equation v= 7- u^2. Putting that into the first equation, u+ 49- 14u^2+ u^4= 11. That is the same as u^4- 14u^2+ u+ 38= 0.

By the 'rational root theorem", the only possible rational solutions to that equation are 1, -1, 2, -2, 19, -19, 38, -38. 2 is the only real rational root - rest of those roots must be irrational or non-real.
 
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