x+y=a z+bx=c bz+xy=d Find yz in terms of a,b,c,d.


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Question Number 54595 by ajfour last updated on 07/Feb/19

$x + y = a$ $z + b x = c$ $b z + x y = d$ $F i n d y z i n t e r m s o f a, b, c, d .$
Commented by mr W last updated on 07/Feb/19

$y = a - x$ $z = c - b x$ $b (c - b x) + x (a - x) = d$ $x^{2} + (b^{2} - a) x + d - b c = 0$ $x = \frac{a - b^{2} \pm \sqrt{{(b^{2} - a)}^{2} - 4 (d - b c)}}{2}$ $y z = (a - x) (c - b x)$ $\Rightarrow y z = \frac{[a + b^{2} \pm \sqrt{{(b^{2} - a)}^{2} - 4 (d - b c)}] [2 c + b^{3} - a b \pm b \sqrt{{(b^{2} - a)}^{2} - 4 (d - b c)}]}{4}$
Commented by ajfour last updated on 07/Feb/19

$T h a n k s S i r, b u t w e c a n m a k e u s e$ $o f e q . i n x^{2} o n l y t o g e t x^{2} i n t e r m s$ $o f x a n d t h e n s u b s t i t u t e, d o i n g t h i s$ $i g e t y z = a c + b^{2} c - b d - (c + b^{3}) x$ $i g e t y z = a c + b^{2} c - b d$ $- (c + b^{3}) \frac{a - b^{2} \pm \sqrt{{(b^{2} - a)}^{2} - 4 (d - b c)}}{2} .$
Commented by mr W last updated on 07/Feb/19

$r i g h t s i r .$
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