SAT Systems with nonlinear equations

Where a line and a curve meet — solve or count intersections.

4% of MathMath · Advanced Math3 question types
~2per test

How to score it

  • Substitute the linear equation into the quadratic, then solve.
  • “Intersect at exactly one point” means the quadratic has one root → discriminant = 0.
  • A line tangent to a parabola touches at exactly one point.

Common traps

  • Keeps an extraneous intersection.
  • “Exactly one point” not recognized as a discriminant condition.
  • Reads only one of two intersection points off a graph.

The 3 question types, with real examples

Line and quadratic intersect

“The graphs intersect at (x, y). What is the value of [y / x²]?”

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From the question bankMedium

y=−x2+9y = - x^{2} + 9 y=2x+6y = 2x + 6 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1+x2x_{1} + x_{2}?

  • A−2-2✓
  • B22
  • C1212
  • D−1-1
Why A

Substituting y=2x+6y = 2x + 6 into the first equation yields 2x+62x + 6 = -x2x^{2} + 9. Rearranging gives x2x^{2} + 2x−3=02x - 3 = 0. By Vieta's formulas, the sum of the roots is -2/1 = -2.

Graph → which solution

“What is the solution (x, y) to this system?”

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From the question bankMedium

In the xy-plane, the graph of y=x2+2x+5y = x^{2} + 2x + 5 intersects the graph of y=6x+1y = 6x + 1 at exactly one point.

What is the solution (x, y) to this system?

  • A(1,8)(1, 8)
  • B(2,13)(2, 13)✓
  • C(−2,13)(-2, 13)
  • D(2,9)(2, 9)
Why B

Settingx2+2x+5=6x+1givesx2−4x+4=0,whichfactorsas(x−2)2=0,sox=2.Substitutingintoy=6x+1givesy=13.Setting x^{2} + 2x + 5 = 6x + 1 gives x^{2} - 4x + 4 = 0, which factors as (x - 2)^{2} = 0, so x = 2. Substituting into y = 6x + 1 gives y = 13.

Identify possible value of x from the system

“[eq1] [eq2]. If (x, y) is a solution to the system, which of the following could be the value of x?”

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From the question bankMedium

x2+y2=25x^{2} + y^{2} = 25 y=x+1y = x + 1 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1y2y_{1} y_{2}?

  • A−12-12✓
  • B1212
  • C−3-3
  • D−11-11
Why A

Substituting x+1x + 1 for y in the first equation yields x2x^{2} + (x+1)2=25(x + 1)^{2} = 25. Expanding gives x2x^{2} + x2x^{2} + 2x+1=252x + 1 = 25, which simplifies to 2x2+2x−24=02x^{2} + 2x - 24 = 0 or x2x^{2} + x−12=0x - 12 = 0. Since y=x+1y = x + 1, the product y1y2y_{1} y_{2} = (x1x_{1} + 1)(x2x_{2} + 1) = x1x2+x1+x2x_{1} x_{2} + x_{1} + x_{2} + 1. By Vieta's formulas, x1x2x_{1} x_{2} = -12 and x1+x2x_{1} + x_{2} = -1, so y1y2y_{1} y_{2} = -12 + (-1) + 1 = -12.

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