Hey dear!!! This is really a very very very long process question..So this question can only be solved using hit and trial method OR trial and error methodHere`s the process√x+y=7x+√y=11√x=7-y√y=11-xthe square root of a number is always the positive square roottherefore y can`t be greater than 7 and x can`t be greater than 11look at the perfect squares 1, 4, 9, 16assuming x and y are perfect squares and a little trial and error...x=9  and y=4
						

							Or if you want a detailed answer here is Or if you want a detailed answer here is it....for understanding some of the processes you have to refer to other webpages mentioned in the ansfor understanding some of the processes you have to refer to other webpages mentioned in the answerGood luck!!!(1) (√x) + y = 7(2) x + √y = 11I did this:Let u = √x and  v = √ySo (1) and (2) can be re-written as:(1`) u + v 2 = 7(2`) u 2 + v = 11(1`) —> u = 7-v 2Substituting into (2`) for u:(7-v 2 )2 + v = 11(49 - 14v 2 + v 4 ) + v = 11v4 - 14v 2 + v + 38 = 0(v-2)(v 3 +2v 2 -10v-19) = 0(v-2)(v-3.13131)(v+1.84813)(v+3.28319) = 0 ( y=4—(from (1))—> √x=3 —> x=9v=3.13131 —> y=9.80510 ——> √x=-2.8051 —> x=7.86859v=-1.84813 —> y=3.41558 ——> √x=3.58442 —> x=12.84807v=-3.28319 —> y=10.77934 ——> √x=-3.77934 —> x=14.28341Check:x=9, y=4:(1) 3+4=7 (ok)(2) 9+2=11 (ok)x=7.86859, y=9.80510(1) -2.8051+9.80510=7 (ok) (only if we take the negative root of x)(2) 7.86859+3.13131=11 (ok)x=12.84807, y=3.41558(1) 3.58442+3.41558=7 (ok)(2) 12.84807-1.84813=11 (ok)(only if we take the negative root of y)x=14.28341, y=10.77934(1)  -3.77934+10.77934=7 (ok)(only if we take the negative root of x)(2)  14.28341-3.28319=11 (ok) (only if we take the negative root of y)Given no other information, √a is defined to be the positive square root of a only —> so only x=9, y=4 is a proper answer