Find all set of ordered triple/s (x,y,z), x,y,z∈ℜ, such that x−y=1−z 3(x^2 −y^2 )=5(1−z^2 ) 7(x^3 −y^3 )=19(1−z^3 ). Please show your solution.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 31286 by 6123 last updated on 05/Mar/18

$F i n d a l l s e t o f o r d e r e d t r i p l e / s (x, y, z), x, y, z \in ℜ, s u c h t h a t$ $x - y = 1 - z$ $3 (x^{2} - y^{2}) = 5 (1 - z^{2})$ $7 (x^{3} - y^{3}) = 19 (1 - z^{3}) .$ $P l e a s e s h o w y o u r s o l u t i o n .$
Commented by 6123 last updated on 06/Mar/18

$T h a n k y o u!!$
	Answered by MJS last updated on 05/Mar/18

	$obviously all triples$ $(x, y, 1) with x = y \in R solve the$ $given equations$ $3 (x^{2} - y^{2}) = 5 (1 - z) (1 + z)$ $from (i) : (1 - z) = (x - y)$ $3 (x - y) (x + y) = 5 (x - y) (1 + z)$ $3 (x + y) = 5 (1 + z)$ $x = \frac{5}{3} (1 + z) - y$ $into (i) : \frac{5}{3} (1 + z) - 2 y = 1 - z$ $y = \frac{1}{3} + \frac{4}{3} z$ $x = \frac{4}{3} + \frac{1}{3} z (from above)$ $into (i i i) :$ $7 ({(\frac{4}{3} + \frac{1}{3} z)}^{3} - {(\frac{1}{3} + \frac{4}{3} z)}^{3}) = 19 - 19 z^{3}$ $z^{3} - \frac{7}{2} z^{2} + \frac{7}{2} z - 1 = 0$ $(z - 1) (z^{2} - \frac{5}{2} z + 1) = 0$ $z_{1} = 1; z_{2} = \frac{1}{2}; z_{3} = 2$ $x_{1} = \frac{5}{3}; x_{2} = \frac{3}{2}; x_{3} = 2$ $y_{1} = \frac{5}{3}; y_{2} = 1; y_{3} = 3$ $so we have$ $(r; r; 1) with r \in R$ $(\frac{3}{2}; 1; \frac{1}{2})$ $(2; 3; 2)$
	Commented by 6123 last updated on 06/Mar/18

	$T h a n k y o u!!!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum