# Solving for multiple unknown variables

Assume for this scenario that:

a = b * c

d = b - c

I only know the value for a and d (in this case 0.075 and 239, respectively), and I need to solve for both b and c. Is this possible to solve given the limited information, and if so what is the best approach to do so?

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This is known as a system of nonlinear simultaneous equations. In this case, it is very easy to solve the system.

First, write $b$ in terms of $d$ and $c$, where $d$ is known:

$$b = d+c.$$

Substitute this into the first equation:

$$a = bc = (d+c)c = cd+c^2$$

Solve this for $c$ using the quadratic formula:

$$c^2+dc-a = 0 \implies c = -\frac{d}{2}\pm \frac{\sqrt{d^2+4a}}{2}$$

Substitute the solutions back into the equation $b = d+c$ to find values of $b$.

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 The problem I am running into is, after solving the quadratic equation I get the results -0.00031 and -239 and I know neither of those to be possible values. Even as positive values, neither of them make sense given the real world number I am trying to solve for. Any idea as to why this would be? – Moses Oct 29 '12 at 19:28 I do not know, because I do not have the context of the problem. All I know is that there are finitely many possible solutions given the input data. – Arkamis Oct 29 '12 at 19:32 My apologies, you were correct in your approach and I was incorrect in my assumptions. Once I adjusted my assumptions your approach produced correct results. Thanks. – Moses Oct 29 '12 at 23:03

Yes, there is a solution. I recommend checking the exact form (not approximate) as provided by wolfram alpha.

Solution

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 Very cool, I did not know wolfram alpha was capable of that kind of analysis. I'll be going back there next time I find myself sans graphic calculator! – Moses Oct 29 '12 at 23:05

Rewrite your two equations: $$\begin{cases} .075 = b\times c \\ 239 = b-c \end{cases}$$ Now solve the second equation for $b$ so that $b=239+c$. Plug this into the first equation in place of $b$: $$.075 = (239+c)\times c$$ Do you know how to solve quadratic equations? Once you have the possible values of $c$, plug those back into our equation for $b$ to get the final answer. This method is called eliminating a variable by substitution.

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