2014 AMC 10A Problems/Problem 21: Difference between revisions
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==Solution== | ==Solution== | ||
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>. | |||
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 
and 
are such that the graphs of 
and 
intersect the 
-axis at the same point. What is the sum of all possible -coordinates of these points of intersection?
Solution
Note that intersects the 
axis at 
, and intersects the -axis at 
. We are given that the 2 graphs intersect the x-axis at the same point, so 
, so 
.
The only possible pairs 
then are 
. These pairs give respective -values of 
which have a sum of 
.
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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