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Sorry for the trivial question. I was wondering if there was an easy way to solve this equation.
The equation is: $$ \frac{30}{15+a} + \frac{30}{15-a} = 4.5 $$ The approach I tried was multiplying both sides by $(15+a)(15-a)$ to get rid of the fractions but I ended up with a really messy quadratic. I am sure there is a simpler way.

Thanks for any help :-)

Edit

Thanks for all the help, it seems there are many ways to do it, you just need a little clever thinking. Just in case anyone's interested I got this equation from an iq test question on a phone app. Which was: 

A motorboat, whose speed is 15km/hr in still water goes 30km downstream    
and comes back in a total of 4 hours 30 minutes. What is the speed    
of the stream's current?

Maybe I was barking up the wrong tree with this approach, the test was only supposed to take 5 minutes and had 10 questions. Is there a simpler way?

c_programmer_jim
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6 Answers6

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Not that messy and a little algebra makes things really simple:

$$\frac{30}{15+a} + \frac{30}{15-a} = 4.5\iff450-30a+450+30a=\frac92(225-a^2)\iff$$

$$\iff1800=2025-9a^2\iff 9a^2=225\iff a^2=25\iff a=\pm 5$$

DonAntonio
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Rewrite the equation as $$ 30\left(\frac{1}{15+a}+\frac{1}{15-a}\right)=\frac{9}{2} $$ or $$ 20\left(\frac{1}{15+a}+\frac{1}{15-a}\right)=3 $$ The term in parentheses can be developed to $$ \frac{(15-a)+(15+a)}{(15+a)(15-a)}=\frac{30}{225-a^2} $$ so your equation becomes $$ 20\frac{30}{225-a^2}=3 $$ or, better, $$ \frac{200}{225-a^2}=1 $$ that is, $$ 200=225-a^2 $$ which should be easy.

Since there are two solutions, it is a quadratic. You can however predict that there will be no first degree term in the final equation because exchanging $a$ with $-a$ doesn't change the equation.

egreg
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Not really. This is a quadratic equation in disguise. You can make the numbers smaller by defining $x=\frac a{15}$. The equation becomes $\frac 2{1+x}+ \frac 2{1-x}=\frac 92$ and when you clear the denominators you get $4(1-x)+4(1+x)=9-9x^2, 0=1-9x^2$ which doesn't seem so messy. Even without the substitution, you won't have a first power term, so can solve the equation just by taking a square root.

Ross Millikan
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If you got a really messy quadratic, then you made a mistake. Remember the identity:

$$ (X+Y)(X-Y)=X^2-Y^2 $$

In particular:

$$ (15+a)(15-a)=225-a^2 $$

If you multiply both sides of your equation by $(15+a)(15-a)$ you get:

\begin{align} 30(15-a)+30(15+a)&=4.5(225-a^2)\\ 2\times30\times15&=4.5(225-a^2)\\ 4.5a^2&=4.5\times225-2\times30\times15\\ a^2&=225-2\times30\times15/4.5=225-900/4.5=225-200=25 \end{align}

This is a quadratic with no linear term, so it should be particularly easy to solve!

John Gowers
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There might be a clever method based on some type of substitution or other insight, but I just did the obvious (without a calculator, by the way): $$\frac{30}{15 + a} \; + \; \frac{30}{15 - a} \; = \; \frac{9}{2}$$ Now divide both sides by $3$ to make the numbers smaller. $$\frac{10}{15 + a} \; + \; \frac{10}{15 - a} \; = \; \frac{3}{2}$$ Now combine the fractions on the left side. $$\frac{10(15 - a) \; + \; 10(15 + a)}{(15 + a)(15 - a)} \; = \; \frac{3}{2}$$ Now cross-multiply. $$20(15 - a) \; + \; 20(15 + a) \; = \; 3(15+a)(15-a)$$ Now expand the left side. $$(20)(15) \; - \; 20a \; + \; (20)(15) \; + \; 20a \; = \; 3(15+a)(15-a)$$ Now cancel the $\;20a$'s $\;$and use the fact that $\;(20)(15) + (20)(15) = (40)(15).$ $$(40)(15) \; = \; 3(15+a)(15-a)$$ Now divide both sides by $3$. $$(40)(5) \; = \; (15+a)(15-a)$$ Now expand the right side. (Really easy because it's a factored form of a difference of squares.) $$(40)(5) \; = \; 15^2 - a^2$$ Now isolate $a^2$. $$a^2 \; = \; 15^2 - (40)(5)$$ Now factor the right hand side. The numbers are small enough that this isn't really needed, but if the numbers were larger, then the method I'll demonstrate could be useful. $$a^2 \; = \; 3^2 \cdot 5^2 \; - \; 8 \cdot 5^2$$ Now factor out $5^2$. $$a^2 \; = \; 5^2 (3^2 - 8)$$ Now do some arithmetic. $$a^2 \; = \; 5^2(9 - 8)$$ $$a^2 \; = \; 5^2$$ Now take the square root of both sides. $$a \; = \; \pm 5$$

Dave L. Renfro
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  • this is sort of the approach I went for but messed up as I haven't done any maths in a while :-) – c_programmer_jim Sep 24 '13 at 20:28
  • I figured as much, which is why I filled in all the steps. By the way, Ross Millikan's method of using $x = \frac{a}{15}$ is the kind of thing I thought might be possible when I wrote my first sentence above, but it's not the kind of thing a typical student at this level should worry about knowing how to do. (I upvoted him for it, however.) – Dave L. Renfro Sep 24 '13 at 20:32
  • ha, I agree. I think that is the sort of maths you can't teach, you just need a bit of foresight and natural ability. I accepted Ross Millikan's answer on it being the most simple, but thanks everyone else for your help :-) – c_programmer_jim Sep 24 '13 at 20:38
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Note this is like the reverse of a partial fraction equation. Working with the fractions $$\frac{30}{15-a}+\frac {30}{15+a} = 30\times \frac {30}{225-a^2}$$ And this makes it clear that you have an equation in $a^2$, rather than in $a$.

Mark Bennet
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