2003 AMC 10B Problem 11

1. An $x$-intercept is where a line reaches height $0$. Both lines start at the point $(10, 15)$, which is at height $15$, so each has to drop $15$ units.
2. Slope $3$ means: every $1$ step to the right, the line goes up $3$. Going left, it goes down $3$ per step. To drop $15$ it needs $15 \div 3 = 5$ steps left. Its $x$-intercept is $10 - 5 = 5$.
3. Slope $5$ means it drops $5$ per step left. To drop $15$ it needs $15 \div 5 = 3$ steps. Its $x$-intercept is $10 - 3 = 7$.
4. The distance between the intercepts is $7 - 5 = 2$.
5. Check with equations: the lines are $y = 3x - 15$ and $y = 5x - 35$. Putting $y = 0$ gives $x = 5$ and $x = 7$.

The answer is $\boxed{\textbf{(A)}\ 2}$.

Slope is "up divided by across," so if you know how far up or down you must go, the distance across is that amount divided by the slope. The steeper line reaches the axis after fewer steps, which is why its intercept is closer to $x = 10$. This avoids writing equations at all.

Both lines pass through $(10, 15)$ and must descend $15$ units to reach the $x$-axis.

On the line of slope $3$, each unit of horizontal movement changes $y$ by $3$, so descending $15$ requires moving $\frac{15}{3} = 5$ units to the left. Its $x$-intercept is $10 - 5 = 5$.

On the line of slope $5$, descending $15$ requires $\frac{15}{5} = 3$ units to the left, so its $x$-intercept is $10 - 3 = 7$.

The distance between the intercepts is $7 - 5 = 2$.

(Equivalently, the lines are $y = 3x - 15$ and $y = 5x - 35$; setting $y = 0$ gives $x = 5$ and $x = 7$.)

The answer is $\boxed{\textbf{(A)}\ 2}$.