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Question Number 97270 by bobhans last updated on 07/Jun/20
solve for all real value of x,y and z giving answer the form (x,y,z) { ((x(x+y)+z(x−y)= 4)),((y(y+z)+x(y−z) = −4)),((z(z+x)+y(z−x) = 5)) :}
solveforallrealvalueofx,yandzgivinganswertheform(x,y,z){x(x+y)+z(x−y)=4y(y+z)+x(y−z)=−4z(z+x)+y(z−x)=5
Commented by john santu last updated on 07/Jun/20
(1) x^2 +xy+xz−yz = 4 (2) y^2 +xy−xz+yz = −4 (3) z^2 −xy+xz+yz = 5 adding (1),(2) ⇒ (x+y)^2 =0 ,x+y = 0 (4) adding (2),(3)⇒ (z+y)^2 =1 ,z+y= ± 1(5) adding (1),(3)⇒ (x+z)^2 = 9, x+z = ± 3 (6) from (4)+(5)+(6) 2x+2y+2z = 4 or 2x+2y+2z = 2 or 2x+2y+2z = −4 2x+2y+2z = −2 { ((x+y+z=2, z = 2 ⇒ { ((x=1, 3)),((y=−1,−3)) :})),((x+y+z=1, z=1 ⇒ { ((x=0,2)),((y=0,−2)) :})),((x+y+z=−2,z=−2⇒ { ((x=−3,−1)),((y=3,1)) :})),((x+y+z=−1,z=−1⇒ { ((x=−2,0)),((y=2,0)) :})) :}
(1)x2+xy+xz−yz=4(2)y2+xy−xz+yz=−4(3)z2−xy+xz+yz=5adding(1),(2)⇒(x+y)2=0,x+y=0(4)adding(2),(3)⇒(z+y)2=1,z+y=±1(5)adding(1),(3)⇒(x+z)2=9,x+z=±3(6)from(4)+(5)+(6)2x+2y+2z=4or2x+2y+2z=2or2x+2y+2z=−42x+2y+2z=−2{x+y+z=2,z=2⇒{x=1,3y=−1,−3x+y+z=1,z=1⇒{x=0,2y=0,−2x+y+z=−2,z=−2⇒{x=−3,−1y=3,1x+y+z=−1,z=−1⇒{x=−2,0y=2,0
Commented by john santu last updated on 07/Jun/20
then the solution is {(1,−1,2),(2,−2,1) ,(−1,1,−2),(−2,2,−1)} (0,0,1),(0,0,−1),(−3,3,−2) ,(3,−3,2) rejected
thenthesolutionis{(1,−1,2),(2,−2,1),(−1,1,−2),(−2,2,−1)}(0,0,1),(0,0,−1),(−3,3,−2),(3,−3,2)rejected
Commented by bobhans last updated on 07/Jun/20
thank you= correct sir
thankyou=correctsir
Answered by 1549442205 last updated on 07/Jun/20
Plus the first eq.and second eq.we get x^2 +2xy+y^2 =0 ⇔(x+y)^2 =0⇔x=−y. Replace into first eq and third eq..we get { ((2xz=4)),((x^2 +z^2 =5)) :}⇔ { (((x+z)^2 =9)),(((x−z)^2 =1)) :}⇔ { ((x+z=±3)),((x−z=±1)) :} ⇔(x;z)∈{(2;−1);(−1;−2);(1;2);(−2;−1)} Therefore,the solutions of given system are: (x;y;z)∈{(2;−2;1);(−1;1;−2);(1;−1;2);(−2;2;−1)}
Plusthefirsteq.andsecondeq.wegetx2+2xy+y2=0⇔(x+y)2=0⇔x=−y.Replaceintofirsteqandthirdeq..weget{2xz=4x2+z2=5⇔{(x+z)2=9(x−z)2=1⇔{x+z=±3x−z=±1⇔(x;z)∈{(2;−1);(−1;−2);(1;2);(−2;−1)}Therefore,thesolutionsofgivensystemare:(x;y;z)∈{(2;−2;1);(−1;1;−2);(1;−1;2);(−2;2;−1)}
Commented by bemath last updated on 07/Jun/20
yess
yess
Commented by bobhans last updated on 07/Jun/20
yess
yess

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