# Difference between revisions of "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

~ dolphin7