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

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Which means that <cmath>-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15</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>.
 
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>.
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It is curious when plugging each ordered pair into the equations, the pair of lines are the same. As an exercise, ask yourself why that is.
 
  
 
==Solution 2==
 
==Solution 2==
  
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First, notice that the value of x cannot exceed 5 because the minimum value for a is 1. Also, notice that for the second equation, it intersects x at <math> 0, -\dfrac{1}{3}, -\dfrac{2}{3}, -1</math> and so on. We then realize that the only integer values for x are <math> -1</math> and <math>-5</math>. We also see that for a fraction to be the value of x, the numerator must divide 5 evenly. So, the only other values are <math>-\dfrac{5}{3}</math> and <math>-\dfrac{1}{3}</math>. Adding, we get <math>\boxed{\textbf{(E)} \: -8}</math>.
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Going off of Solution 1, for the first equation, notice that the value of <math>x</math> cannot be less than <math>-5</math>. We also know for the first equation that the values of <math>x</math> have to be <math>5</math> divided by something. Also, for the second equation, the values of <math>x</math> can only be <math>-\frac13,-\frac23,-\frac33, \dots</math>. Therefore, we see that, the only values common between the two sequences are <math>-1, -5, -\frac13,-\frac53</math>, and adding them up, we get for our answer, <math>\boxed{\textbf{(E)} \: -8}</math>.
  
 
==See Also==
 
==See Also==
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{{MAA Notice}}
 
{{MAA Notice}}
  
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[[Category: Introductory Algebra Problems]]
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[[Category: Advanced Algebra Problems]]

Revision as of 17:55, 21 October 2021

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 1

Note that when $y=0$, the $x$ values of the equations should be equal by the problem statement. We have that \[0 = ax + 5 \implies x = -\dfrac{5}{a}\] \[0 = 3x+b \implies x= -\dfrac{b}{3}\] Which means that \[-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15\] 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 $\boxed{\textbf{(E)} \: -8}$.

Solution 2

Going off of Solution 1, for the first equation, notice that the value of $x$ cannot be less than $-5$. We also know for the first equation that the values of $x$ have to be $5$ divided by something. Also, for the second equation, the values of $x$ can only be $-\frac13,-\frac23,-\frac33, \dots$. Therefore, we see that, the only values common between the two sequences are $-1, -5, -\frac13,-\frac53$, and adding them up, we get for our answer, $\boxed{\textbf{(E)} \: -8}$.

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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