• Mar 10th 2011, 03:23 PM
sirellwood
Can anybody solve these two equations with two unknowns for me?

$\displaystyle \frac{a}{a+b}$= $\displaystyle \frac{4}{102}$, and

$\displaystyle \frac{a*b}{(a+b)^2(a+b+1)}$ = 0.0003658

Basically I am doing a Statistics question about beta distribution and this has cropped up during the question, and when i get my answers, i keep gettin 'a' as a negative number (which is not permitted in beta distribution)

Thanks
• Mar 10th 2011, 04:12 PM
HallsofIvy
Quote:

Originally Posted by sirellwood
Can anybody solve these two equations with two unknowns for me?

$\displaystyle \frac{a}{a+b}$= $\displaystyle \frac{4}{102}$, and

$\displaystyle \frac{a*b}{(a+b)^2(a+b+1)}$ = 0.0003658

Basically I am doing a Statistics question about beta distribution and this has cropped up during the question, and when i get my answers, i keep gettin 'a' as a negative number (which is not permitted in beta distribution)

Thanks

From $\displaystyle \frac{a}{a+ b}= \frac{4}{102}= \frac{2}{51}$
$\displaystyle 51a= 2(a+ b)= 2a+ 2b$
$\displaystyle 49a= 2b$
$\displaystyle b= \frac{49}{2}a$

Replace the "b" in $\displaystyle \frac{ab}{(a+ b)(a+ b+ 1}= 0.0003658$
with that and you will have a cubic equation for a.
• Mar 10th 2011, 04:19 PM
sirellwood
thanks, yeah i got a=4 and b=98 (3 s.f) better answers!
• Mar 10th 2011, 04:26 PM
Soroban
Hello, sirellwood!

Quote:

Can anybody solve these two equations with two unknowns for me?

. . $\displaystyle \dfrac{a}{a+b}\:=\:\dfrac{4}{102} \qquad \dfrac{ab}{(a+b)^2(a+b+1)} \:=\: 0.0003658$

Basically I am doing a Statistics question about beta distribution
and this has cropped up during the question.
And when i get my answers, i keep gettin 'a' as a negative number. ??

First, I'll re-write the equations:

. . [1] .$\displaystyle \dfrac{a}{a+b} \:=\:\dfrac{2}{51}$

. . [2] .$\displaystyle \dfrac{ab}{(a+b)^2(a+b+1)} \:=\:d\;\text{ (for decimal)}$

From [1], we have: .$\displaystyle \dfrac{a}{a+b} \:=\:\dfrac{2}{51} \quad\Rightarrow\quad b \:=\:\dfrac{49}{2}a$ .[3]

From [2], we have:

. . $\displaystyle \displaystyle \underbrace{\left(\frac{a}{a+b}\right)}_{\text{Thi s is }\frac{2}{51}}\cdot \frac{b}{(a+b)(a+b+1)} \:=\:d$

. . . . . . $\displaystyle \displaystyle \frac{2}{51}\cdot\frac{b}{(a+b)(a+b+1)} \:=\:d$

. . . . . . . . . . . . . . . . . . . . $\displaystyle 2b \;=\;51d(a+b)(a+b+1)$

Substitute [3]:

. . $\displaystyle 2\left(\frac{49}{2}a\right) \;=\;51d\left(a + \frac{49}{2}a\right)\left(a + \frac{49}{2}a + 1\right)$

. . . . .$\displaystyle 49a \;=\;51d\left(\frac{51}{2}a\right)\left(\frac{51a+ 2}{2}\right)$

$\displaystyle \text{Divide by }a\!:\;\;49 \:=\:\dfrac{2601d(51a+2)}{4} \quad\Rightarrow\quad 196 \;=\;132,\!651ad + 5202d$

. . . . . $\displaystyle 132,\!651ad \;=\;196 - 5202d \quad\Rightarrow\quad a \;=\;\dfrac{196-5202d}{132,\!651d}$

Subsstitute $\displaystyle d = 0.0003658$

. . $\displaystyle a \;=\;\dfrac{196 - 5202(0.0003658)}{132,\!651(0.0003658)} \;=\;4.000044621$

Therefore: .$\displaystyle a\:\approx\;4$

Edit: Too slow again . . . *sigh
.