### Instruction: Two lines have the same non-zero $y$-intercept. The first line has a slope of 10 and an $x$-intercept of $(s, 0)$. The second line has a slope of 6 and an $x$-intercept of $(t, 0)$. What is the ratio of $s $ to $t$? Express your answer as a common fraction.### Response: I know that the equation of a line in slope-intercept form is $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept.
Since the two lines have the same $y$-intercept, I can call it $b$ for both of them.
Then the equation of the first line is $y = 10x + b$ and the equation of the second line is $y = 6x + b$.
To find the $x$-intercepts, I need to set $y$ equal to zero and solve for $x$.
For the first line, I get $0 = 10x + b$, so $x = -\frac{b}{10}$.
That means $s = -\frac{b}{10}$.
For the second line, I get $0 = 6x + b$, so $x = -\frac{b}{6}$.
That means $t = -\frac{b}{6}$.
To find the ratio of $s$ to $t$, I need to divide $s$ by $t$ and simplify.
I get $\frac{s}{t} = \frac{-\frac{b}{10}}{-\frac{b}{6}} = \frac{6}{10} = \frac{3}{5}$.