# How do I solve for $6$ variables with only $3$ equations?

I have 3 equations that need to be satisfied:

$$\displaystyle \frac{27}{30} = \frac{R_2}{R_1 + R_2}\\$$

$$\displaystyle \frac{15}{30} = \frac{R_4}{R_3 + R_4}\\$$

$$\displaystyle \frac{3}{30} = \frac{R_6}{R_5 + R_6}$$

Any values of $R_1-R_6$ that satisfy the above $3$ questions will work for me, but they need to be between $10000 - 100000$.

What is the easiest way to solve the values for $R_1-R_6$?

• Do you want to define all solutions or just a solution? – JB King Feb 20 '13 at 19:43

If any solution works take for $R_3=R_4$ anything you like, For \begin{align*} R_6&=10000\\ R_5&=90000\\ R_2&=90000\\ R_1&=10000\\ R_3&=50000\\ R_4&=50000\\ \end{align*}

A general solutation is \begin{align*} R_3&=R_4\\ R_5&=9 \cdot R_6\\ R_2&= 9 \cdot R_1\\ \end{align*}

Take reciprocals of each equation, e.g. $$\frac{30}{27}=\frac{R_1+R_2}{R_2}=1+\frac{R_1}{R_2}\implies R_1=\left(\frac{30}{27}-1\right)R_2=\frac1{9}R_2.$$

You are free to choose $R_2,R_4, R_6$ and then find $R_1, R_3, R_5$ by these equations.

• That + on the rightmost side must be a minus... – DonAntonio Feb 20 '13 at 19:45

$$I\;\;\;\;\;\;\;\frac{r_2}{r_1+r_2}=\frac{9}{10}\Longleftrightarrow10r_2=9r_1+9r_2\Longleftrightarrow r_2=9r_1$$

so you can take $\,r_1=10,000\,\,,\,\,r_2=90,000\,$

Do something similar for the other equations