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Squaring First equ both side (Under root x+y)^2=7^2 then we get x+y=49 then y=49-x -----Equ. No.----(3) Squaring Second equ both side ^2=11^2 we get x^2+y=121 putting the value y from (3) equ. X^2+(49-X)=121 after solving we get X^2-X+49-121=0 then X^2-X-72=0 Using factorization splitting...

read more Squaring First equ both side (Under root x+y)^2=7^2 then we get x+y=49 then y=49-x -----Equ. No.----(3) Squaring Second equ both side [x+(under root y)]^2=11^2 we get x^2+y=121 putting the value y from (3) equ. X^2+(49-X)=121 after solving we get X^2-X+49-121=0 then X^2-X-72=0 Using factorization splitting middle term we get X^2-9X+8X-72=0 then X(X-9)+8(x-9)=0 then (x-9)(x-8)=0 then (X-9)=0 or (X+8)=0 Hence X=9 or X=-8 Answer read less

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Since we have ?x and a y in one equation and ?y and a x in the other, x and y must be positive(because u cannot hav a negative value inside the Sq Root), Since ?x + y = 7, then y ? 7 Since x + ?y = 11, then x ? 11 Assuming x and y are perfect squares(because u cannot hav a negative value inside...

read more Since we have ?x and a y in one equation and ?y and a x in the other, x and y must be positive(because u cannot hav a negative value inside the Sq Root), Since ?x + y = 7, then y ? 7 Since x + ?y = 11, then x ? 11 Assuming x and y are perfect squares(because u cannot hav a negative value inside the Sq Root), we try the following: x = 0, 1, 4, 9 y = 0, 1, 4 The only values that work are x = 9, y = 4 ?x + y = ?9 + 4 = 3 + 4 = 7 x + ?y = 9 + ?4 = 9 + 2 = 11 Implies x=9 and y=4 Please correct me if there is something wrong. read less

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x+?y=11... . . ?x+y= 7....... . from equation ?y=11--x .so y= 121--22x+x^2 we can substititute this value of y in equation ?x +121 --22x+x^2=7 ?x = --121+-22x--x^2+7 solving irt by trial & error method if WE HAVE x==9 THEN, WE HAVE LHS=?x==?9==3.............. & RHS=-121...

read more x+?y=11... .[.1 ] . ?x+y= 7.......[2] . from equation [1] ?y=11--x .so y= 121--22x+x^2 we can substititute this value of y in equation[2] ?x +121 --22x+x^2=7 ?x = --121+-22x--x^2+7 solving irt by trial & error method if WE HAVE x==9 THEN, WE HAVE LHS=?x==?9==3..............[A] & RHS=-121 +[22X9]--81= -121 +198 -81= 3...[B] COMPARING BOTH QUANTITITIES WE FIND LHS=RHS HENCE x=9 is the correct solution Now we can have y=7--?x= 7--?9=7--3=4 hence the required value 0f x=9 & y=4 read less

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By simple substitution we can find x = 9 and y = 4

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x+?y=11... . . ?x+y= 7....... . from equation ?y=11--x .so y= 121--22x+x^2 we can substititute this value of y in equation ?x +121 --22x+x^2=7 ?x = --121+-22x--x^2+7 solving irt by trial & error method if WE HAVE x==9 THEN, WE HAVE LHS=?x==?9==3.............. & RHS=-121...

read more x+?y=11... .[.1 ] . ?x+y= 7.......[2] . from equation [1] ?y=11--x .so y= 121--22x+x^2 we can substititute this value of y in equation[2] ?x +121 --22x+x^2=7 ?x = --121+-22x--x^2+7 solving irt by trial & error method if WE HAVE x==9 THEN, WE HAVE LHS=?x==?9==3..............[A] & RHS=-121 +[22X9]--81= -121 +198 -81= 3...[B] COMPARING BOTH QUANTITITIES WE FIND LHS=RHS HENCE x=9 is the correct solution Now we can have y=7--?x= 7--?9=7--3=4 hence the required value 0f x=9 & y=4 read less

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X=9 y=4

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y = 4 and x = 9

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