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Linear-Quadratic Systems on the SAT

SAT Math · Advanced Math · Updated October 2026

These systems come up in the second module, as plain equations and as graphs. Desmos makes the plain ones quick: graph both and click the crossing points. Questions with a constant like $c$ or $k$ still need the algebra.

How to solve them

Solve a system with a line and a parabola by substitution. Set the two expressions for $y$ equal to each other, move everything to one side, and solve the quadratic. Each $x$ you find gives one solution point, and you put it back into the line to get its $y$.

A line and a parabola can cross twice, touch once, or not meet at all. If a question asks how many solutions there are, look at the discriminant $b^2 - 4ac$ of the combined quadratic. Positive means two, zero means one, and negative means none.

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The parabola $y = x^2$ and the line $y = x + 6$ cross at $(-2, 4)$ and $(3, 9)$. Each crossing point is a solution of the system.

Question types

You seeDo this
$y = x^2 + \ldots$ and $y = mx + b$Set them equal, move everything to one side, factor or use the quadratic formula.
"How many solutions?"Find the discriminant of the combined quadratic, or count the crossings in Desmos.
"Exactly one solution", with a constantThe line is tangent: set the discriminant equal to $0$ and solve for the constant.
A horizontal line $y = k$It meets an upward parabola twice above the vertex, once at the vertex, and never below it.
"Which point is a solution?"The point must satisfy both equations. Test each choice in both.

Worked examples

Example 1: two crossing points

Solve $y = x^2$ and $y = x + 6$.

  1. Set equal: $x^2 = x + 6$, so $x^2 - x - 6 = 0$.
  2. Factor: $(x - 3)(x + 2) = 0$, so $x = 3$ or $x = -2$.
  3. Put each into the line: $(3, 9)$ and $(-2, 4)$. Those are the two crossing points in the figure.

Example 2: a tangent line

For what value of $c$ do $y = x^2 + c$ and $y = 2x$ have exactly one solution?

  1. Set equal: $x^2 - 2x + c = 0$.
  2. One solution means the discriminant is $0$: $(-2)^2 - 4(1)(c) = 0$.
  3. $4 - 4c = 0$, so $c = 1$. The line just touches the parabola at $(1, 2)$.

Example 3: a horizontal line

How many solutions do $y = x^2 - 4x + 7$ and $y = 3$ have?

  1. Set equal: $x^2 - 4x + 4 = 0$, which is $(x - 2)^2 = 0$.
  2. Only $x = 2$ works, so there is exactly one solution.
  3. That makes sense: the vertex of the parabola is $(2, 3)$, and the line $y = 3$ touches it right there.

Where students lose points

Forgetting the second solution. A quadratic usually gives two $x$ values. Each one is a separate crossing point.

Finding $x$ and stopping. Remember that a solution is a point, with both an $x$ and a $y$. If the question asks for $y$, put $x$ back into the line.

Using the parabola's discriminant alone. Combine the two equations first. The discriminant that matters is of the combined quadratic.

Testing a point in only one equation. $(2, 4)$ is on $y = x^2$, but that doesn't make it a solution of the system.

Sign slips when moving terms. $x^2 = x + 6$ becomes $x^2 - x - 6 = 0$.

Practice

Try these on paper, and check the plain ones in Desmos afterward.

Easy
How many solutions does the system $y = x^2$ and $y = 4$ have?
  • AZero
  • BExactly one
  • CExactly two
  • DInfinitely many
Show solution
Answer: C. $x^2 = 4$ gives $x = 2$ or $x = -2$, so the solutions are $(2, 4)$ and $(-2, 4)$.
EasyStudent-produced response
The system $y = x^2 - 1$ and $y = 3$ has a solution $(x, y)$ with $x > 0$. What is the value of $x$?
Enter your answer.
Show solution
Answer: $2$. $x^2 - 1 = 3$ gives $x^2 = 4$, so $x = 2$ or $x = -2$. The positive value is $2$.
Medium
Which point is a solution to the system $y = x^2$ and $y = x + 6$?
  • A$(3, 9)$
  • B$(2, 4)$
  • C$(3, 6)$
  • D$(-3, 3)$
Show solution
Answer: A. $(3, 9)$ works in both: $9 = 3^2$ and $9 = 3 + 6$. Choice B is on the parabola but not the line, since $2 + 6 \ne 4$.
MediumStudent-produced response
The system $y = x^2 + 2x$ and $y = 2x + 9$ has two solutions, $(x_1, y_1)$ and $(x_2, y_2)$. What is the value of $y_1 + y_2$?
Enter your answer.
Show solution
Answer: $18$. $x^2 + 2x = 2x + 9$ gives $x^2 = 9$, so $x = 3$ or $x = -3$. The $y$-values are $15$ and $3$, which add to $18$.
Medium
How many solutions does the system $y = x^2 - 4x + 7$ and $y = 3$ have?
  • AZero
  • BExactly two
  • CInfinitely many
  • DExactly one
Show solution
Answer: D. $x^2 - 4x + 4 = 0$ is $(x - 2)^2 = 0$, so only $x = 2$ works. The line touches the parabola at its vertex $(2, 3)$.
MediumStudent-produced response
The system $y = -x^2 + 6$ and $y = x$ has two solutions. What is the product of their $x$-coordinates?
Enter your answer.
Show solution
Answer: $-6$. $-x^2 + 6 = x$ gives $x^2 + x - 6 = 0$, so $(x + 3)(x - 2) = 0$. The $x$-coordinates are $-3$ and $2$, and their product is $-6$.
HardStudent-produced response
In the system $y = x^2 + c$ and $y = 2x$, $c$ is a constant. For what value of $c$ does the system have exactly one solution?
Enter your answer.
Show solution
Answer: $1$. $x^2 - 2x + c = 0$ has one solution when the discriminant is $0$: $4 - 4c = 0$, so $c = 1$.
Hard
In the system $y = x^2$ and $y = k$, $k$ is a constant. For which values of $k$ does the system have no real solutions?
  • A$k = 0$
  • B$k > 0$
  • C$k < 0$
  • D$k = 1$
Show solution
Answer: C. $x^2 = k$ has no real solutions when $k$ is negative, because a square is never negative. The line $y = k$ is then below the vertex of the parabola. With $k = 0$ there is one solution.
Hard
The system $y = 2x^2 - 3x - 2$ and $y = x + 4$ has a solution $(a, b)$ with $a > 0$. What is the value of $b$?
  • A$7$
  • B$3$
  • C$-1$
  • D$13$
Show solution
Answer: A. $2x^2 - 3x - 2 = x + 4$ gives $2x^2 - 4x - 6 = 0$, or $x^2 - 2x - 3 = 0$. So $x = 3$ or $x = -1$. With $a = 3$, $b = 3 + 4 = 7$. Choice B is the $y$-value for $x = -1$.
HardStudent-produced response
In the system $y = x^2 - 6x + k$ and $y = -1$, $k$ is a constant. For what value of $k$ does the system have exactly one solution?
Enter your answer.
Show solution
Answer: $8$. $x^2 - 6x + k + 1 = 0$ has one solution when $36 - 4(k + 1) = 0$, so $k + 1 = 9$ and $k = 8$. Then the vertex of $y = x^2 - 6x + 8$ is $(3, -1)$, right on the line.

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