Systems of Nonlinear Equations

When a Line Meets a Curve

A nonlinear system pairs at least one equation that is not a straight line — usually a parabola like $y = x^2 - 3$ — with another equation. A solution is still an ordered pair $(x, y)$ that satisfies both equations, and graphically it is still an intersection point. The new wrinkle is that a line and a parabola can meet twice, once, or not at all, so these questions test both your algebra and your ability to reason about how many solutions exist.

The Reliable Method: Substitution

Substitution almost always beats elimination here, because one equation is usually already solved for $y$. The plan is fixed: set the two expressions equal, move everything to one side, and solve the resulting quadratic.

Solve the system $y = x^2 - 2x - 1$ and $y = 2x - 4$.

Step 1 — set the expressions equal. Both equal $y$, so $x^2 - 2x - 1 = 2x - 4$.

Step 2 — collect on one side. Subtract $2x$ and add $4$:

$x^2 - 4x + 3 = 0.$

Step 3 — factor. Two numbers that multiply to $3$ and add to $-4$ are $-1$ and $-3$: $(x - 1)(x - 3) = 0$, so $x = 1$ or $x = 3$.

Step 4 — find $y$ for each $x$. Use the simpler line $y = 2x - 4$: at $x = 1$, $y = -2$; at $x = 3$, $y = 2$.

The solutions are $(1, -2)$ and $(3, 2)$ — two intersection points. Always substitute your $x$-values back into an original equation to recover $y$: the answer is a pair of points, not just the $x$-values.

Counting Intersections With the Discriminant

Sometimes the SAT does not want the solutions — it wants to know how many there are, or the value of a constant that produces a specific count. After substitution you have a quadratic $ax^2 + bx + c = 0$, and its discriminant $b^2 - 4ac$ counts the real solutions without your ever solving it:

Discriminant $b^2 - 4ac$	Real solutions	Geometry
Positive	Two	Line cuts the curve twice
Zero	One	Line is tangent (touches once)
Negative	None	Line misses the curve entirely

For $x^2 - 4x + 3 = 0$ above, the discriminant is $(-4)^2 - 4(1)(3) = 16 - 12 = 4 > 0$ — confirming the two intersection points we found.

Tangency: The Discriminant-Equals-Zero Move

The most common "hard" version asks for a value that makes a line tangent to a parabola. Tangent means the curves touch at exactly one point, which means the quadratic has exactly one solution, which means the discriminant is zero. For instance, to make $y = k$ tangent to $y = x^2 - 6x + 11$, substitute to get $x^2 - 6x + (11 - k) = 0$ and set the discriminant to zero:

$\begin{aligned} (-6)^2 - 4(1)(11 - k) &= 0 \\ 36 - 44 + 4k &= 0 \\ 4k &= 8 \\ k &= 2 \end{aligned}$

So the horizontal line $y = 2$ just grazes the parabola at its vertex. Translating "tangent" into "discriminant $= 0$" is the single insight these items reward.

When There Is No Real Solution

If substitution yields a quadratic with a negative discriminant, the graphs never meet — the line passes entirely above or below the parabola. This is a legitimate SAT answer. Consider $y = x^2 + 1$ and $y = x - 2$: setting them equal gives $x^2 - x + 3 = 0$, whose discriminant is $(-1)^2 - 4(1)(3) = 1 - 12 = -11 < 0$. There are no real intersection points, so the system has no real solution.

Reading the Graph

A quick sketch confirms your algebra: a line drawn across a parabola will visibly cross it in two places, brush it at one place, or clear it entirely. The picture and the discriminant always agree, so use one to check the other under time pressure.