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I have two equations:

\begin{align*} 3 \ln x + \ln y &= 3 \\ 4 \ln x - 6 \ln y &= -7 \\ \end{align*}

Do I just proceed as I have learned with adding equations resulting in:

\begin{align*} 18 \ln x + 6 \ln y &= 18 \\ 4 \ln x - 6 \ln y &= -7 \\ 22 \ln x &= 11 \\ \end{align*}

Then solve from there?

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Yes, that is exactly what you do. I assume that going from $22\ln x =11$ to $x=e^{1/2}$ is obvious for you. However, the way you express your calculation will almost certainly lose you marks. For example, $18\ln x +6\ln y=18$. –  André Nicolas Dec 13 '11 at 16:30
    
Thanks, that answered my question exactly. Your correction is the way it's written down, I just made the error here on the site. –  erimar77 Dec 13 '11 at 16:33

1 Answer 1

up vote 3 down vote accepted

Yes, that is how the system is handled. In a more complicated situation, we might want to make a substitution, letting $u=\ln x$ and $v=\ln y$. Then our two equations become $$3u +v=3, \qquad 4u-6v=-7.$$ This is a system of two linear equations in the two unknowns $u$ and $v$.

If we have a system of quite a few linear equations in quite a few unknowns, we need to adopt a systematic approach. For our simple system, almost anything sensible will work. For example, we can try to "eliminate" $v$ (that was your strategy). Multiply each side of the first equation by $6$. We obtain $18u+3v=18$.

From the equations $18u+3v=18$ and $4u-6v=-7$, we obtain, by adding, that $22u=11$, so $u=1/2$. Substituting this in one of the equations, we get $v=3/2$.

So $\ln x=1/2$ and $\ln y=3/2$. It follows that $x=e^{1/2}$ and $y=e^{3/2}$.

Comment: As noted above, for this simple system almost any reasonable strategy will work. For example, in high school one might be encouraged to rewrite the first equation as $v=3-3u$, and then substitute for $v$ in the second equation. We would get $$4u-6(3-3u)=-7,$$ which, after some manipulation, yields $22u=11$.

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