Test 8 Section 4 #27 please

Yeah, this is a pretty tricky one! The key to getting through it is recognizing that whenever you have a point on a line, you can plug the coordinates into the line’s equation. Because the question wants to know about \dfrac{r}{p}, we should start by getting our equations in terms of r and p.

We know (p, r) is on y = x + b, so we substitute:

    \begin{align*}y&=x+b\\r&=p+b\end{align}

Likewise, we know that (2p, 5r) is on y = 2x + b, so we substitute:

    \begin{align*}y&=2x+b\\5r&=2(2p)+b\\5r&=4p+b\end{align}

From there, we can use the fact that both equations have a b in them to our advantage. Solve each for b, then set them equal to each other:

    \begin{align*}5r&=4p+b\\5r-4p&=b\\\\r&=p+b\\r-p&=b\\\\5r-4p&=r-p\end{align}

Now just combine like terms and solve for \dfrac{r}{p}:

    \begin{align*}5r-4p&=r-p\\4r&=3p\\r&=\dfrac{3}{4}p\\\dfrac{r}{p}&=\dfrac{3}{4}\end{align}

 

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