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Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y) =7 Then X , Y = ?

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If we use the proper method, the solution is very lengthy and also it involves polynomial to the power 4. However, if we use hit and trial method, it becomes very easy. sqrt x + y =11, sqrt y + x =7, it implies that x and y are perfect squares less than 11.There are three perfect squares less than...
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If we use the proper method, the solution is very lengthy and also it involves polynomial to the power 4. However, if we use hit and trial method, it becomes very easy. sqrt x + y =11, sqrt y + x =7, it implies that x and y are perfect squares less than 11.There are three perfect squares less than 11. They are 1,4,9. We put the values of each and find the answer. The answer is x=4 and y =9. Take another example- sqrt x +y =51 , x+sqrt y= 11. Here, x and y are perfect squares less than 51. So, it can be 1,4,9,16,25,36,49. By hit and trial, we find that x =4 and y=49. Hope this helps. Thanks and regards Smita read less
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(Rootx}+y=11 and x+{rooty)=7 (Rootx)=11-y and rooty=7-x Rootx=11-(7-x)^2 x=^2 x =121-22(7-x)^2+7-x)^4 Let7-x=t t^ 4-22t+t+114=0 t=3is one proper root So t=7-x=3 x=4 Soy=9
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MSc.,Mathematics

?x + y = 7 x + ?y = 11 Let ?x = a and ?y = b a + b² = 7 ---------- (1) a² + b = 11 ---------- (2) (1) × a² and (2) × a a³ + a²b² = 7a² ---------- (1) a³ + ab = 11a ---------- (2) subtract a²b² -- ab = 7a² -- 11a ab(ab -- 1) = a(7a -- 11) b(ab -- 1) = (7a -- 11) ab² -- b = 7a...
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?x + y = 7 x + ?y = 11 Let ?x = a and ?y = b a + b² = 7 ---------- (1) a² + b = 11 ---------- (2) (1) × a² and (2) × a a³ + a²b² = 7a² ---------- (1) a³ + ab = 11a ---------- (2) subtract a²b² – ab = 7a² – 11a ab(ab – 1) = a(7a – 11) b(ab – 1) = (7a – 11) ab² – b = 7a – 11 – b = 7a – ab² – 11 – b = a(7 – b²) – 11 a = (11 – b)/(7 – b²) Let b = 1, 2, ... when b = 2 a = (11 – b)/(7 – b²) = (11 – 2)/(7 – 4) = 9/3 = 3 so a = 3 and b = 2 ?x = 3 and ?y = 2 x = 9 and y = 4 read less
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solution of above question: root(x) +y=11 ....(i) x + root(y) =7 ....(ii) from (i) and (ii) we have, (11 -y)^2 + root(y) =7 put root(y) =t in above then eq. reduces to polynomial of degree 4 , i.e. t^4 -22*t^2 +t +114 =0 =>(t-3)(t^3 +3*t^2 -13*t -38 *t)=0 ........using...
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solution of above question: root(x) +y=11 ....(i) x + root(y) =7 ....(ii) from (i) and (ii) we have, (11 -y)^2 + root(y) =7 put root(y) =t in above then eq. reduces to polynomial of degree 4 , i.e. t^4 -22*t^2 +t +114 =0 =>(t-3)(t^3 +3*t^2 -13*t -38 *t)=0 ........using remainder theorem a factor of above eq. is t-3 hence t=3 so y=9 and hence root(x) =2 answer is x=4,y=9 read less
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Teacher

x=4,y=9
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Headstart to assured 95+ score in math

X=4, y=9. Better use hit and trial method.
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Math magician

X=4 , Y=9
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Tutor

4,9
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Teacher

x=4, y=9
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Trainer

4 and 9
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