Difference between revisions of "2014 AMC 10A Problems/Problem 21"
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==Solution== | ==Solution== | ||
− | Note that <math>y= | + | Note that when <math>y=0</math>, the <math>x</math> values of the equations should be equal by the problem statement. We have that |
+ | <cmath>0 = ax + 5 \implies x = -\dfrac{5}{a}</cmath> <cmath>0 = 3x+b \implies x= -\dfrac{b}{3}</cmath> | ||
+ | Which means that <cmath>-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15</cmath> | ||
+ | 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>. | ||
− | + | --happiface. | |
==See Also== | ==See Also== |
Revision as of 00:21, 8 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 when , the
values of the equations should be equal by the problem statement. We have that
Which means that
The only possible pairs
then are
. These pairs give respective
-values of
which have a sum of
.
--happiface.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.