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AoPS Wiki talk:Problem of the Day/August 7, 2011

Revision as of 12:04, 7 August 2011 by Kingofmath101 (talk | contribs) (August 7, 2011, Problem of the Day- Solution)
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Solution:

First, notice that the second and third equations have identical numerators such that they are not equal to $1$. So, divide both sides of the second equation by $2$, and divide both sides of the third equation by $3$. Then our system transforms:

$\frac{1}{x + y} = \frac{1}{7}$

$\frac{2}{y + z} = \frac{2}{9}$

$\frac{3}{x + z} = \frac{3}{8}$

$\Downarrow$

$\frac{1}{x + y} = \frac{1}{7}$

$\frac{1}{y + z} = \frac{1}{9}$

$\frac{1}{x + z} = \frac{1}{8}$

Now, we make the assumption that:

$x + y \neq 0$

$y + z \neq 0$

$x + z \neq 0$

This is because we are now going to take the inverse ($-1$st power) of all three equations: $\Downarrow$

$x + y = 7$

$y + z = 9$

$x + z = 8$

We now have this system of equations to work with. Start by adding all three equations together, which gives us

$2x + 2y + 2z = 24$

Divide both sides of the equation by $2$:

$x + y + z = 12$

Knowing this makes it very easy to find the values of the variables from here. First subtract the first equation in our linear polynomial system from this equation, which gives us $z = 5$. We now have the equation

$y + 5 = 9 \Rightarrow y = 4$.

We can put this value off $y$ into the first equation of our system:

$x + 4 = 7$

Subtracting $4$ from both sides of this equation gives us $x = 3$.

Acknowledgements Art of Problem Solving's Intermediate Algebra and my former MATHCOUNTS instructor taught me some techniques that I used in this solution.

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