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I am stuck solving for $a-ar^2=112$ and $ar-ar^3=84$

I got $a=\frac{112}{1-r^2}$ and $a=\frac{84}{r-r^3}$

Then I got a cubic equation. But answer only has 1 value for a & r, so I think there must be an easier way?

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2  
Try to divide second equation by first. –  Norbert Nov 19 '11 at 10:05
    
$a(1-r^2)=112$ and $ar(1-r^2)=84$ –  pedja Nov 19 '11 at 10:07
    
Your cubic equation $112(r-r^3)=84(1-r^2)$ or $112r(1+r)(1-r)=84(1+r)(1-r)$ has the solutions $r=1,-1,\frac{3}{4}$, but the first two of these do not fit the conditions of the question. –  Henry Nov 19 '11 at 10:20

1 Answer 1

up vote 2 down vote accepted

$a-ar^2=112\iff a(1-r^2)=112\qquad \qquad(1)$

$ar-ar^3=84\iff ar(1-r^2)=84\qquad\qquad(2)$

Since $r=1$ and $r=-1$ are not solutions of the equations, the equation on the right in (1) is equivalent to

$ a={112\over 1-r^2}. $

Now, substituting this into the equation on the right of (2):

$ {112\over 1-r^2}\cdot r (1-r^2)=84\iff112 r =84\iff r={84\over112}={3\over4}$.

Going back to the first equation

$a(1-(3/4)^2)=112\Rightarrow a={112\over 1-{9\over16}}={112\over 7/16}={16\cdot 112\over 7}=16^2=256$.

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How did you do the "Solve for a" part? isit: $a=\frac{112}{1-r^2}$. Then $\frac{112}{1-r^2} \cdot r (1-r^2)=84$. Then it still gives a cubic equation? –  Jiew Meng Nov 19 '11 at 11:15
    
Cancel the $1-r^2$ terms. You'll wind up with an equivalent equation since $r=-1$ and $r=1$ aren't solutions (as seen looking at $a-ar^2=112$) –  David Mitra Nov 19 '11 at 11:24
    
Oh man... how did I get so blur :( ... –  Jiew Meng Nov 19 '11 at 11:24

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