2014 AMC 10A Problem 21

Set $y = 0$ in each equation. The line $y = ax + 5$ meets the $x$-axis at $x = -\frac{5}{a}$, and $y = 3x + b$ meets it at $x = -\frac{b}{3}$. These are equal exactly when
$$\frac{5}{a} = \frac{b}{3} \quad\Longleftrightarrow\quad ab = 15.$$

Since $a$ and $b$ are positive integers, $(a, b)$ is one of $(1, 15),\ (3, 5),\ (5, 3),\ (15, 1)$. The common $x$-intercept is $-\frac{5}{a}$, giving
$$-5,\quad -\frac{5}{3},\quad -1,\quad -\frac{1}{3}.$$

Their sum is
$$-\left(5 + 1 + \frac{5}{3} + \frac{1}{3}\right) = -(6 + 2) = -8.$$

The answer is $\boxed{\textbf{(E)}\ {-8}}$.

"Same $x$-intercept" is one equation in $a$ and $b$; with the integer restriction it becomes a factoring problem, $ab = 15$, whose solutions are just divisor pairs. The intercepts themselves are then determined by $a$ alone, so list the divisors of $15$ and add. Recognizing when a coordinate-geometry condition collapses to a Diophantine equation is the whole skill here.

Use $x = -\frac{b}{3}$ instead: the possible $b$ are the divisors $1, 3, 5, 15$ of $15$, and the sum of intercepts is $-\frac{1 + 3 + 5 + 15}{3} = -\frac{24}{3} = -8$. This is a quick consistency check on the arithmetic.

• Adding only the integer intercepts, or summing the values of a or b instead of the intercepts -5/a.
• Missing the divisor pairs $(1, 15)$ and $(15, 1)$, which gives $-\frac{5}{3} - 1 = -\frac{8}{3}$, or allowing negative divisors even though $a, b$ are positive.