Albert1
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ab=1
bc=2
cd=3
de=4
ea=5
find a,b,c,d,e
bc=2
cd=3
de=4
ea=5
find a,b,c,d,e
anemone said:My solution:
[TABLE="class: grid, width: 800"]
[TR]
[TD]From $ab=1$ and $bc=2$, we have:
$2ab=bc$
$2ab-bc=0$
$b(2a-c)=0$
Since $b \ne 0$, $2a-c=0$ must be true or $c=2a$.[/TD]
[TD]From $cd=3$ and $c=2a$, we have:
$(2a)d=3$
$2ad=3$[/TD]
[TD]From $de=4$ and $ea=5$ and $2ad=3$, we have:
$ade^2=4(5)$
$2ad(e^2)=2(20)$
$3(e^2)=40$
$\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$[/TD]
[TD]$\begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*}$
$\begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*}$
$\begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*}$
$\begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}$
[/TD]
[/TR]
[/TABLE]
kaliprasad said:I thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others
kaliprasad said:As I do not know how to put square root I have put power 1/2
kaliprasad said:Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$
Similarly you can find the rest
good solution (Clapping)kaliprasad said:As I do not know how to put square root I have put power 1/2
Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$
Similarly you can find the rest