# 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

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 $0, -\dfrac{1}{3}, -\dfrac{2}{3}, -1$ and so on. We then realize that the only integer values for x are $-1$ and $-5$. 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 $-\dfrac{5}{3}$ and $-\dfrac{1}{3}$. Adding, we get $\boxed{\textbf{(E)} \: -8}$.