Question 144·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Type $x^2+y=10$ and $y^2+x=10$ on separate Desmos lines. An intersection makes both equations true. Click the upper-right and lower-left crossings; Desmos labels them about $(2.702,2.702)$ and $(-3.702,-3.702)$. These have $x=y$, but don't stop there: the curves also enter the mixed-sign regions.

Click the two mixed-sign intersections. Desmos labels them about $(3.541,-2.541)$ and $(-2.541,3.541)$. Swapping the coordinates gives a different ordered pair, so these are two more points to count. Each distinct point shared by both graphs counts as one solution. The system has $4$ ordered pairs. Choice D.

Both equations equal $10$, so subtract the second from the first to cancel their right sides:

Remove the parentheses and collect terms:

Use the difference of squares $x^2-y^2=(x-y)(x+y)$, and write $-x+y$ using the same factor:

Factor out $x-y$:

By the zero-product rule, at least one factor must be zero. So either $x=y$ or $x+y=1$.

For $x=y$, replace $y$ with $x$ in the first equation:

Subtract $10$:

The graph of this quadratic opens upward. It is below zero at $x=0$ and rises on both ends, so it crosses zero twice. Each of those two real values of $x$ gives one pair $(x,x)$.

For $x+y=1$, subtract $x$:

Replace $y$ in the first equation:

Remove the parentheses:

Subtract $10$ and collect constants:

This quadratic also opens upward and is below zero at $x=0$, so it has two real roots. Each gives one pair through $y=1-x$.

If a pair had both $x=y$ and $x+y=1$, it would have $x=y=\frac12$. But $0<x<1$ makes $x^2<x$, so $x^2+y<x+y=1$, not $10$. The cases share no solution, so their two pairs each give $2+2=4$ ordered pairs. Choice D.