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

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

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

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

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

$\frac{a*b}{(a+b)^2(a+b+1)}$ = 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 $\frac{a}{a+ b}= \frac{4}{102}= \frac{2}{51}$
$51a= 2(a+ b)= 2a+ 2b$
$49a= 2b$
$b= \frac{49}{2}a$

Replace the "b" in $\frac{ab}{(a+ b)(a+ b+ 1}= 0.0003658$
with that and you will have a cubic equation for a.

3. thanks, yeah i got a=4 and b=98 (3 s.f) better answers!

4. Hello, sirellwood!

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

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

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

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

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

From [2], we have:

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

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

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

Substitute [3]:

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

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

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

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

Subsstitute $d = 0.0003658$

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

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

Edit: Too slow again . . . *sigh
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