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Question Number 155991 by cortano last updated on 07/Oct/21

  { ((a(x+2)+y=3a)),((a+2x^3 =y^3 +(a+2)x^3 )) :}   solve for x &y in term a

{a(x+2)+y=3aa+2x3=y3+(a+2)x3 solveforx&yinterma

Commented byRasheed.Sindhi last updated on 07/Oct/21

Mr cortano, if you want answer  in terms of a , that is x,y depepend  upon a the following answer is  right the answer you are looking   for:  x=((−1−2a^2 ±(√(12a^2 −3)))/(2−2a^2 ))  y=a−a(((−1−2a^2 ±(√(12a^2 −3)))/(2−2a^2 )))  Fix value of a,calculate x & y  there is the solution which is  satisfied by both given equations  for that fixed value of a.  For easy cases substitute a=±13  x=((−1−2(±13)^2 ±(√(12(±13)^2 −3)))/(2−2(±13)^2 ))      =((−339±45)/(−336))=(7/8),(8/7)  y_1 =13−13((7/8))=((13)/8)  y_2 =13−13((8/7))=−((13)/7)  (x,y)=((7/8),((13)/8)),((8/7),−((13)/7)) for a=13  VERIFICATION:    { ((a(x+2)+y=3a)),((a+2x^3 =y^3 +(a+2)x^3 )) :}     { ((13((7/8)+2)+((13)/8)=3(13))),((13+2((7/8))^3 =(((13)/8))^3 +(13+2)((7/8))^3 )) :}     { ((13(((7+16)/8))+((13)/8)=3(13))),((((13.8^3 +2.7^3 )/8^3 )=((13^3 +15.7^3 )/8^3 ))) :}      { ((13(((23+1)/8))=3(13)⇒13(3)=3(13))),((((7342)/8^3 )=((7342)/8^3 ))) :}     V  E  R  I  F  I  E  ^(              ) D

Mrcortano,ifyouwantanswer intermsofa,thatisx,ydepepend uponathefollowingansweris righttheansweryouarelooking for: x=12a2±12a2322a2 y=aa(12a2±12a2322a2) Fixvalueofa,calculatex&y thereisthesolutionwhichis satisfiedbybothgivenequations forthatfixedvalueofa. Foreasycasessubstitutea=±13 x=12(±13)2±12(±13)2322(±13)2 =339±45336=78,87 y1=1313(78)=138 y2=1313(87)=137 (x,y)=(78,138),(87,137)fora=13 VERIFICATION: {a(x+2)+y=3aa+2x3=y3+(a+2)x3 {13(78+2)+138=3(13)13+2(78)3=(138)3+(13+2)(78)3 {13(7+168)+138=3(13)13.83+2.7383=133+15.7383 {13(23+18)=3(13)13(3)=3(13)734283=734283 VERIFIED

Answered by Rasheed.Sindhi last updated on 07/Oct/21

  { ((a(x+2)+y=3a.......(i))),((a+2x^3 =y^3 +(a+2)x^3 ......(ii))) :}   solve for x &y in term a  (i):y=3a−a(x+2)=a(1−x)...(iii)  (ii):y^3 =a+2x^3 −(a+2)x^3 =a+x^3 (2−a−2)      =a(1−x^3 )..........(iv)  (iv)&(iii):a(1−x^3 )=a^3 (1−x)^3   1−x^3 =a^2 (1−x)^3     [a≠0]  1−x^3 −a^2 (1−x)^3 =0  (1−x)(1+x+x^2 )−a^2 (1−x)^3 =0  (1+x+x^2 )−a^2 (1−x)^2 =0  [x≠1]  x^2 +x+1−a^2 (x^2 −2x+1)=0  x^2 −a^2 x^2 +x+2xa^2 +1−a^2 =0  (1−a^2 )x^2 +(1+2a^2 )x+(1−a^2 )=0  x=((−(1+2a^2 )±(√((1+2a^2 )^2 −4(1−a^2 )^2 )))/(2(1−a^2 )))    △=1+4a^4 +4a^2 −4(1−2a^2 +a^4 )          =1+4a^4 +4a^2 −4+8a^2 −4a^4           =12a^2 −3  x=((−1−2a^2 ±(√(12a^2 −3)))/(2−2a^2 ))  (iii):y=a(1−x)     y=a−a(((−1−2a^2 ±(√(12a^2 −3)))/(2−2a^2 )))               (to be simplified)

{a(x+2)+y=3a.......(i)a+2x3=y3+(a+2)x3......(ii) solveforx&yinterma (i):y=3aa(x+2)=a(1x)...(iii) (ii):y3=a+2x3(a+2)x3=a+x3(2a2) =a(1x3)..........(iv) (iv)&(iii):a(1x3)=a3(1x)3 1x3=a2(1x)3[a0] 1x3a2(1x)3=0 (1x)(1+x+x2)a2(1x)3=0 (1+x+x2)a2(1x)2=0[x1] x2+x+1a2(x22x+1)=0 x2a2x2+x+2xa2+1a2=0 (1a2)x2+(1+2a2)x+(1a2)=0 x=(1+2a2)±(1+2a2)24(1a2)22(1a2) =1+4a4+4a24(12a2+a4) =1+4a4+4a24+8a24a4 =12a23 x=12a2±12a2322a2 (iii):y=a(1x) y=aa(12a2±12a2322a2) (tobesimplified)

Commented bycortano last updated on 07/Oct/21

x=1 solution sir⇒y=0

x=1solutionsiry=0

Answered by Rasheed.Sindhi last updated on 07/Oct/21

  { ((a(x+2)+y=3a)),((a+2x^3 =y^3 +(a+2)x^3 )) :}   solve for x &y in term a   { ((y=3a−a(x+2)=a(1−x)...A)),((y^3 =a+2x^3 −(a+2)x^3 =a(1−x^3 )...B)) :}   A & B :  a^3 (1−x)^3 =a(1−x^3 )   a^2 (1−x)^3 −(1−x^3 )=0  (1−x){a^2 (1−x)−(1+x+x^2 )}=0  x=1 ∣ a^2 −a^2 x−1−x−x^2 =0^★   If x=1  a(x+2)+y=3a⇒a(1+2)+y=3a  ⇒3a+y=3a⇒y=0  x=1,y=0 (Answer free of a)    ^★ Will produce results in terms of a

{a(x+2)+y=3aa+2x3=y3+(a+2)x3 solveforx&yinterma {y=3aa(x+2)=a(1x)...Ay3=a+2x3(a+2)x3=a(1x3)...B A&B:a3(1x)3=a(1x3) a2(1x)3(1x3)=0 (1x){a2(1x)(1+x+x2)}=0 x=1a2a2x1xx2=0 Ifx=1 a(x+2)+y=3aa(1+2)+y=3a 3a+y=3ay=0 x=1,y=0(Answerfreeofa) Willproduceresultsintermsofa

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