Difference between revisions of "2014 AMC 10A Problems/Problem 21"

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==Solution==
 
==Solution==
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Note that <math>y=ax+5</math> intersects the <math>x-</math>axis at <math>(-\frac{5}{a}, 0)</math>, and <math>y=3x+b</math> intersects the <math>x</math>-axis at <math>(-\frac{b}{3}, 0)</math>. We are given that the 2 graphs intersect the x-axis at the same point, so <math>-\frac{5}{a}=-\frac{3}{b}</math>, so <math>ab=15</math>.
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The only possible pairs <math>(a,b)</math> then are <math>(a,b) = (1,15), (3,5), (5,3), (15, 1)</math>. These pairs give respective <math>x</math>-values of <math>-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}</math> which have a sum of <math>\boxed{\textbf{(E)} \: -8}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 17:53, 7 February 2014

Problem

Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$-axis at the same point. What is the sum of all possible $x$-coordinates of these points of intersection?

$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8}$

Solution

Note that $y=ax+5$ intersects the $x-$axis at $(-\frac{5}{a}, 0)$, and $y=3x+b$ intersects the $x$-axis at $(-\frac{b}{3}, 0)$. We are given that the 2 graphs intersect the x-axis at the same point, so $-\frac{5}{a}=-\frac{3}{b}$, so $ab=15$.

The only possible pairs $(a,b)$ then are $(a,b) = (1,15), (3,5), (5,3), (15, 1)$. These pairs give respective $x$-values of $-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}$ which have a sum of $\boxed{\textbf{(E)} \: -8}$.

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 10 Problems and Solutions

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