Suppose q and p are linearly related such that p = 30 when q = 5, and p = 50 when q = 7. Find p when q = 15.

Steps to Solve

1. Find the slope

Since $p$ and $q$ are linearly related, we can express the relationship as $p = mq + b$, where $m$ is the slope and $b$ is the y-intercept.

We are given two points: $(5, 30)$ and $(7, 50)$. We can use these points to find the slope $m$.

$m = \frac{p_2 - p_1}{q_2 - q_1} = \frac{50 - 30}{7 - 5} = \frac{20}{2} = 10$

2. Find the y-intercept

Now that we have the slope $m = 10$, we can use one of the points to find the y-intercept $b$. Let's use the point $(5, 30)$.

$30 = 10(5) + b$ $30 = 50 + b$ $b = 30 - 50 = -20$

3. Write the linear equation

Now we have the slope $m = 10$ and the y-intercept $b = -20$. We can write the linear equation as:

$p = 10q - 20$

4. Find p when q = 15

Substitute $q = 15$ into the equation to find the value of $p$.

$p = 10(15) - 20$ $p = 150 - 20$ $p = 130$