Question 193·Hard·Algebra

If the point on $p$ is 3 higher than the point on $q$ for the same $x$, write an equation relating $y_p$ and $y_q$.

After substitution, you should get one equation in $x$ with fractions. Clear denominators or combine like terms carefully.

Enter the two equations in the standard $x$-$y$ plane:

$y=-\frac{5}{8}x+\frac{11}{2}$

$y=x-\frac{3}{2}$

Because $y_p=y_q+3$, graph the line that is 3 units above $q$:

$y=\left(x-\frac{3}{2}\right)+3$

(which simplifies to $y=x+\frac{3}{2}$).

Find the intersection point of the graphs from Step 1 (line $p$) and Step 2 (the shifted line). The $x$-coordinate of that intersection is the requested value.

From the graph, line $p$ passes through $A\,(0,\tfrac{11}{2})$ and $B\,(8,\tfrac{1}{2})$.

Its slope is

Since $p$ crosses the $y$-axis at $\tfrac{11}{2}$, an equation for line $p$ is

From the graph, line $q$ passes through $C\,(0,-\tfrac{3}{2})$ and $D\,(6,\tfrac{9}{2})$.

Its slope is

Since $q$ crosses the $y$-axis at $-\tfrac{3}{2}$, an equation for line $q$ is

“The y-coordinate on $p$ is 3 greater than the y-coordinate on $q$ (at the same $x$)” means

Substitute the two expressions:

Simplify the right side and solve:

Multiply both sides by $\tfrac{8}{13}$:

So the correct choice is $\frac{32}{13}$.