Question 7·25 Super-Hard SAT Advanced Math Questions·Advanced Math

Type $x^2-3y=12$ exactly as given. Desmos draws a parabola, a U-shaped curve. Click its lowest point to see $(0,-4)$. Keep that lower part of the graph in view when you look for crossings.

Add $y^2-3x=12$ and click every place the curves meet. Desmos shows points near $(-3.791,0.791)$, $(-2.275,-2.275)$, $(0.791,-3.791)$, and $(5.275,5.275)$. Each shared point is one ordered pair, even though it lies on both curves. There are four such pairs. Choice B.

Both equations equal $12$, so subtract the second from the first to get an equation equal to zero:

Remove the parentheses:

Use the difference of squares, $x^2-y^2=(x-y)(x+y)$, then factor out the shared $x-y$:

By the zero-product property, a product equals zero if at least one factor equals zero. So don't divide by $x-y$; that would lose a case:

Rearrange each case:

For the $x=y$ case, put $x$ in place of $y$ in the first original equation:

Each real value of $x$ here gives one ordered pair $(x,x)$.

Add $(\frac32)^2=\frac94$ to both sides to make a perfect square:

Rewrite the left side:

A square can equal this positive number in two ways, so this case gives two pairs.

For $x+y=-3$, write $y=-3-x$ and substitute into the first original equation:

Combine terms:

Add $\frac94$ to both sides:

Rewrite the left side:

The positive right side gives two more $x$-values, each with one $y=-3-x$. Neither has $x=-\frac32$, the only value that could make $x=y$ and $x+y=-3$ at once. So the cases don't overlap: the system has four ordered pairs. Choice B.