Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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(xy)=4y(y+z)+x(yz)=4z(z+x)+y(zx)=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+xzyz=4(2)y2+xyxz+yz=4(3)z2xy+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=0x=y.Replaceintofirsteqandthirdeq..weget{2xz=4x2+z2=5{(x+z)2=9(xz)2=1{x+z=±3xz=±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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com