2014 AMC 10A Problems/Problem 21

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, $\boxed{\textbf{(E)} \: -8}$ .

Solution 3

Similar to the above solutions except we're using the equations $a = -\frac{-5}{x}$ and $b = -3x$ With this, we know that c has to be negative. Doing some math, we find $x$ to be $-1, -5, -\frac{1}{3},$ and $-\frac{5}{3}$

Adding them up gives you our answer: $\boxed{\textbf{(E)} \: -8}$ .

~Starshooter11

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc10a/375

~ dolphin7

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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