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Session Length
~17 min
Adaptive Checks
15 questions
Transfer Probes
15
Nonlinear systems on the SAT involve solving two equations simultaneously where at least one equation is not linear. The most common configuration pairs a linear equation with a quadratic equation, though circle-line and rational-function intersections also appear. These problems fall under the Advanced Math domain of the SAT and test your ability to use substitution to reduce a system to a single equation in one variable, then apply quadratic techniques to find solutions.
The key algebraic tool is substitution: replace $y$ (or $x$) from the linear equation into the nonlinear equation, then simplify to a standard quadratic $ax^2 + bx + c = 0$. From there, factoring, the quadratic formula, or discriminant analysis determines the number and values of solutions. The SAT frequently asks how many intersection points exist, what the sum or product of the $x$-coordinates is (Vieta's formulas), or for what parameter value a line is tangent to a curve (discriminant equals zero).
Graphically, these problems correspond to finding where a line crosses or touches a parabola, circle, or other curve. A positive discriminant means two intersection points, a zero discriminant means the line is tangent (one point), and a negative discriminant means no real intersections. Developing fluency with these connections between algebra and geometry is essential for the hardest SAT Math questions and builds a foundation for calculus and analytic geometry.
One step at a time.
Replace one variable using the linear equation and substitute into the nonlinear equation to produce a single equation in one variable, typically a quadratic.
Example: Given $y = 2x + 1$ and $y = x^2$, substitute: $x^2 = 2x + 1 \Rightarrow x^2 - 2x - 1 = 0$.
For $ax^2 + bx + c = 0$, the discriminant $D = b^2 - 4ac$ determines the number of real solutions: $D > 0$ gives two, $D = 0$ gives one, $D < 0$ gives none.
Example: For $x^2 - 4x + 4 = 0$: $D = 16 - 16 = 0$, so there is exactly one solution ($x = 2$).
Setting a linear equation equal to a quadratic produces a quadratic equation whose solutions are the $x$-coordinates of intersection points. A tangent line gives $D = 0$.
Example: For $y = x^2$ and $y = 2x + k$: $x^2 - 2x - k = 0$. Tangency when $D = 4 + 4k = 0$, so $k = -1$.
For $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$: sum $= -\frac{b}{a}$ and product $= \frac{c}{a}$. These allow answering SAT questions without fully solving.
Example: For $x^2 + x - 6 = 0$: sum of roots $= -1$, product of roots $= -6$.
Substituting a linear equation into $x^2 + y^2 = r^2$ yields a quadratic in one variable. The discriminant determines 0, 1, or 2 intersection points.
Example: $x^2 + y^2 = 25$ with $y = 4$: $x^2 + 16 = 25 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$ (two points).
A line is tangent to a parabola when the resulting quadratic has exactly one solution, meaning $D = 0$. Solving $D = 0$ for the parameter finds the tangency condition.
Example: For $y = x^2$ and $y = -x + k$: $x^2 + x - k = 0$, $D = 1 + 4k = 0$, so $k = -\frac{1}{4}$.
The SAT often asks 'how many solutions' a system has. After substitution, use the discriminant: $D > 0$ yields 2, $D = 0$ yields 1, $D < 0$ yields 0.
Example: For $y = x^2 + 1$ and $y = -2$: $x^2 + 1 = -2 \Rightarrow x^2 = -3$. No real solutions ($D < 0$).
A parabola $y = (x - h)^2 + k$ has its minimum (or maximum) at $y = k$. A horizontal line $y = c$ intersects at one point when $c = k$ and at two points when $c > k$ (for upward-opening).
Example: $y = (x - 1)^2$ has vertex at $(1, 0)$. The line $y = 0$ touches it at exactly one point; $y = 4$ crosses at two points.
More terms are available in the glossary.
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