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Question Number 35892 by behi83417@gmail.com last updated on 25/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18

$i a m v e r y n e a r t o f i n d t h e a n s w e r . . . i t s a v e r y g o o d$ $p r o b l e m . . .$
Commented by Rasheed.Sindhi last updated on 29/May/18

${We}^{'} re waiting for your answer sir .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/May/18

	$t = \frac{- 1}{y} . . . . (1)$ $p u t t i n g v a l u e o f t i n (t + y + x z = 0)$ $y + x z - \frac{1}{y} = 0 . . . (2)$ $x + z = 10 . . . . (3)$ $p u t t i n g v a l u e i n (x t + y x = 6)$ $x . \frac{- 1}{y} + y z = 6 . . . . (4)$ $p u t t i n g z = 10 - x i n (y + x z - \frac{1}{y} = 0)$ $y + x (10 - x) - \frac{1}{y} = 0$ $y^{2} + 10 x y - x^{2} y - 1 = 0$ $p u t t i n g z = 10 - x i n (\frac{- x}{y} + y z = 6)$ $\frac{- x}{y} + y (10 - x) = 6$ $- x + 10 y^{2} - {x y}^{2} = 6 y$ $x (1 + y^{2}) = 10 y^{2} - 6 y$ $x = \frac{10 y^{2} - 6 y}{1 + y^{2}}$ $y^{2} + 10 x y - x^{2} y - 1 = 0$ $y^{2} + 10 y (\frac{10 y^{2} - 6 y}{1 + y^{2}}) = 1 + y {(\frac{10 y^{2} - 6 y}{1 + y^{2}})}^{2}$ $\frac{y^{2} + y^{4} + 100 y^{3} - 60 y^{2}}{1 + y^{2}} = 1 + y \frac{(100 y^{4} - 120 y^{3} + 36 y^{2}}{1 + 2 y^{2} + y^{4}}$ $d i t t o l h s = \frac{1 + 2 y^{2} + y^{4} + 100 y^{5} - 120 y^{4} + 36 y^{3}}{{(1 + y^{2})}^{2}}$ $(1 + y_{}^{2}) (y^{2} + y^{4} + 100 y^{3} - 60 y^{2}) =$ $1 + 2 y^{2} + y^{4} + 100 y^{5} - 120 y^{4} + 36 y^{3}$ $y^{2} + y^{4} + 100 y^{3} - 60 y^{2} + y^{4} + y^{6} + 100 y^{5} - 60 y^{4} =$ $1 + 2 y^{2} + 36 y^{3} - 119 y^{4} + 100 y^{5} = y^{6} + 100 y^{5} - 58 y^{4}$ $+ 100 y^{3} - 59 y^{2}$ $c o n t d$ $y^{6} - 58 y^{4} + 119 y^{4} + 100 y^{3} - 36 y^{3} - 59 y^{2} - 2 y^{2} - 1 = 0$ $y^{6} + 61 y^{4} + 64 y^{3} - 61 y^{2} - 1 = 0$ $y^{3} + 61 y + 64 - \frac{61}{y} - \frac{1}{y^{3}} = 0$ $(y^{3} - \frac{1}{y^{3}}) + 61 (y - \frac{1}{y}) + 64 = 0$ $k = y - \frac{1}{y}$ $k^{3} = y^{3} - 3 y^{2} . \frac{1}{y} + 3 y . \frac{1}{y^{2}} - \frac{1}{y^{3}}$ $k^{3} = y^{3} - \frac{1}{y^{3}} - 3 (y - \frac{1}{y})$ $k^{3} + 3 k = y^{3} - \frac{1}{y^{3}}$ $(k^{3} + 3 k) + 61 k + 64 = 0$ $k^{3} + 64 k + 64 = 0$ $k (k^{2} + 64) + 64 = 0$ $i h a v e s o m e t h i n g t o s a y a b o u t s o l v a b i l i t y o f t h e$ $e q u t i o n$ $1) k s h o u l d h a v e (- v e) v a l u e$ $2) b u t k^{2} + 64 i s + v e s o k^{2} + 64 > 64$ $3) s o k (k^{2} + 64) c a n n e v e r b e e q u a l t o (- 64)$ $t o m a k e t h e e q u a t i o n f e a s i b l e$ $s o c l n c l u s i o n k^{3} + 64 k + 64 c a n n o t b e z e r o$ $i h a v e t o t r y t o s o l v e a n o t h e r w a y . .$ $c o n t d$
	Commented by behi83417@gmail.com last updated on 26/May/18

	$t h a n k y o u v e r y m u c h s i r f o r y o u r w o r k i n g .$ $i a m w a i t i n g f o r y o u r f i n a l a n s w e r .$
	Commented by behi83417@gmail.com last updated on 28/May/18

	$d e a r m r t a n m a y! i t h i n k o n e f a c t o r o f$ $y o u r e q u a t i o n i s : y^{2} + y - 1$ $a n d e q u a t i o n : k^{3} + 64 k + 64 = 0 a t l e a s t$ $h a v e o n r o o t t h a t a p p e r o x i m a l l y e q u a i l s$ $t o : - 1 (k = - . 9824)$
	Commented by tanmay.chaudhury50@gmail.com last updated on 29/May/18

	$t h a n x$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18

	$l e t y = a t = \frac{- 1}{a}$ $t + y = a - \frac{1}{a}$ $t + y = \frac{(a + 1) (a - 1)}{a}$ $s o x z = - \frac{(a + 1) (a - 1)}{a}$ $s i n c e t + y + x z = 0$ $x z = \frac{(1 + a) (1 - a)}{a}$ $l e t$ $x = \frac{1 - a}{a} z = 1 + a$ $x = \frac{1 + a}{a} z = 1 - a$ $x = 1 - a z = \frac{1 + a}{a}$ $x + z = 10$ $1 + a + \frac{1 - a}{a} = 10$ $a + a^{2} + 1 - a = 10 a$ $a^{2} - 10 a + 1 = 0$ $a = \frac{10 \pm \sqrt{100 - 4}}{2}$ $a = \frac{10 \pm 4 \sqrt{6}}{2}$ $a = 5 + 2 \sqrt{6} a n d 5 - 2 \sqrt{6}$ $p l s w a i t$
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