Problem 430

Suppose the slope of the three parallel lines is $m$.

The equation of a line with slope $m$ and $y$-intercept 2 is $y=mx+2$.

To find the $x$-intercept in terms of $m$, we set $y=0$ and solve for $x$.

Doing this, we obtain $mx+2=0$ or $x = -\dfrac{2}{m}$.

Similarly, the line with slope $m$ and $y$-intercept $3$ has $x$-intercept $-\dfrac{3}{m}$.

Also, the line with slope $m$ and $y$-intercept $4$ has $x$-intercept $-\dfrac{4}{m}$.

Since the sum of the $x$-intercepts of these lines is $36$, then $\left(-\dfrac{2}{m}\right)+\left(-\dfrac{3}{m}\right)+\left(-\dfrac{4}{m}\right) = 36$.

Multiplying both sides by $m$, we obtain $-2-3-4=36m$ and so $36m=-9$ or $m = -\dfrac{1}{4}$.