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How do you solve a system that pairs a linear equation with a nonlinear one, such as a line and a parabola?

Systems of equations in two variables with a nonlinear equation: solve a line-and-parabola system by substitution, interpret the number of intersection points, and use the discriminant to count solutions.

A focused answer to the Digital SAT Advanced Math skill of solving systems with a nonlinear equation: substituting a line into a parabola, finding intersection points, and using the discriminant to count how many solutions a system has.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this skill is asking
  2. The substitution method
  3. A worked substitution
  4. Counting intersections with the discriminant
  5. When to graph instead

What this skill is asking

A nonlinear system pairs a linear equation with a nonlinear one (most often a line and a parabola) and asks for the points that satisfy both. Geometrically, those are the intersection points of the two graphs. The Digital SAT (Advanced Math domain) tests solving such systems by substitution and reasoning about how many intersection points there are, frequently via the discriminant.

The substitution method

One equation is already a function, which makes substitution natural.

A worked substitution

The combined equation is the whole game.

Counting intersections with the discriminant

A line and a parabola can cross twice, touch once, or miss. After substitution you get a quadratic, and its discriminant tells you which case applies, exactly as for any quadratic: a positive discriminant gives two intersection points, a zero discriminant gives one (the line is tangent to the parabola), and a negative discriminant gives none. The SAT uses this for "for what value of kk is the line tangent to the curve" questions: form the combined quadratic in terms of kk, set its discriminant to zero, and solve for kk. This translates a geometric tangency condition into a clean algebraic equation.

When to graph instead

Substitution is the rigorous method, but on the Digital SAT, graphing in Desmos is often faster for "how many solutions" or "find the intersection" questions: type both equations and count or click the crossing points. Graphing is especially handy when the algebra would be messy or when you only need the number of solutions rather than their exact values. Reserve the algebraic substitution for questions that ask for an exact coordinate, an equation involving a parameter, or a tangency condition that needs the discriminant, where reading a graph alone would not give the exact value or the parameter relationship.

Two graphing curves can also be a parabola and a parabola or a line and a circle, not only a line and a parabola, and the same logic applies: substitute to eliminate one variable and solve the resulting one-variable equation, or graph both and count intersections. A line and a circle, for instance, can meet at two points (the line is a chord), one point (the line is tangent), or none (the line misses the circle), exactly mirroring the line-and-parabola cases. When a system mixes a linear and a quadratic relationship in any guise, the strategy does not change: reduce to one variable by substitution, solve, and interpret the count of real solutions as the count of intersection points. Holding this single template in mind keeps every nonlinear system, whatever shapes it involves, from feeling unfamiliar.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Digital SAT Math (style)1 marksAt how many points do y=x2y = x^2 and y=x+6y = x + 6 intersect? (A) 00 (B) 11 (C) 22 (D) 33
Show worked answer β†’

The correct answer is (C), 2.

Set the expressions equal: x2=x+6x^2 = x + 6, so x2βˆ’xβˆ’6=0x^2 - x - 6 = 0, which factors as (xβˆ’3)(x+2)=0(x - 3)(x + 2) = 0. The solutions x=3x = 3 and x=βˆ’2x = -2 are two distinct values, so the line and the parabola intersect at two points: (3,9)(3, 9) and (βˆ’2,4)(-2, 4).

Digital SAT Math (style)1 marksThe system {y=x2+2y=k\begin{cases} y = x^2 + 2 \\ y = k \end{cases} has exactly one solution. What is the value of kk? (A) 00 (B) 11 (C) 22 (D) 44
Show worked answer β†’

The correct answer is (C), 2.

Substitute: x2+2=kx^2 + 2 = k, so x2=kβˆ’2x^2 = k - 2. This has exactly one solution when kβˆ’2=0k - 2 = 0 (giving x=0x = 0), so k=2k = 2. Geometrically, the horizontal line y=2y = 2 just touches the parabola's vertex (0,2)(0, 2).

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