Systems of Linear and Nonlinear Equations

Introduction

A system of a linear and a nonlinear equation pairs a line with a curve — most often a parabola — and asks you to find the points where they meet. On the SAT, these questions appear in the Advanced Math domain (≈35% of the section) and test whether you can substitute strategically, produce and solve a quadratic, and connect algebra to the graphical picture of intersections. You may be asked for the coordinates of the solutions, the number of solutions, or a specific coordinate value.

Core Concept

What "solving the system" means geometrically

Every solution $(x, y)$ of the system lies on both graphs at once — it is an intersection point. A line can cross a parabola at 0, 1, or 2 points.

Discriminant of the resulting quadratic	Intersections
$\Delta > 0$	2 real solutions
$\Delta = 0$	1 real solution (line is tangent to parabola)
$\Delta < 0$	0 real solutions (no intersection)

The substitution strategy

Given $y = mx + b$ (line) and $y = ax^2 + bx + c$ (parabola):

Substitute the expression for $y$ from the line into the parabola equation. Now there is only one variable.
Collect all terms on one side to get a quadratic set equal to zero.
Solve by factoring or the quadratic formula.
Back-substitute each $x$ -value into the linear equation to find the matching $y$ -value.
Verify each ordered pair in the original parabola equation.

Key Formulas & Rules

Substitution: Replace $y$ in $y = ax^2 + bx + c$ with the linear expression $mx + b_0$ :

mx + b_0 = ax^2 + bx + c \implies ax^2 + (b - m)x + (c - b_0) = 0

Quadratic formula (must be memorized — not on the reference sheet):

x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}

where $Ax^2 + Bx + C = 0$ .

Discriminant (determines the number of intersections):

\Delta = B^2 - 4AC

Vieta's formulas (useful shortcut when the question only asks for a sum or product of the $x$ -values):

x_1 + x_2 = -\frac{B}{A}, \qquad x_1 x_2 = \frac{C}{A}

These apply after the system is reduced to $Ax^2 + Bx + C = 0$ .

Note: The discriminant $\Delta = B^2 - 4AC$ appears in the sibling note "The Discriminant and Number of Solutions," which treats it in depth for single quadratics. Here we apply it specifically to determine how many times a line and a parabola cross.

Worked Examples

Example 1

The system below has two solutions.

y = 4x - 3 \qquad y = x^2 + 2x - 3

What is the sum of the $y$ -coordinates of the two solutions?

A) $-3$ B) $2$ C) $5$ D) $8$

Step 1 — Substitute. Since both expressions equal $y$ , set them equal:

4x - 3 = x^2 + 2x - 3

Step 2 — Collect terms.

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

Step 3 — Factor.

x(x - 2) = 0 \implies x = 0 \text{ or } x = 2

Step 4 — Find the $y$ -values using the linear equation $y = 4x - 3$ :

$x = 0$ : $\;y = 4(0) - 3 = -3$ . Check in parabola: $0 + 0 - 3 = -3$ ✓
$x = 2$ : $\;y = 4(2) - 3 = 5$ . Check in parabola: $4 + 4 - 3 = 5$ ✓

Step 5 — Answer the question. Sum of $y$ -coordinates: $-3 + 5 = \mathbf{2}$ .

Why the distractors fail:

A) $-3$ : Records only the $y$ -coordinate of the first solution, stopping after $x = 0$ .
C) $5$ : Records only the $y$ -coordinate of the second solution, stopping after $x = 2$ .
D) $8$ : Replaces $-3$ with $3$ (ignores the negative sign) and adds: $3 + 5 = 8$ .

Answer: B) $2$

Desmos check

Desmos check: Graph $y = 4x - 3$ and $y = x^2 + 2x - 3$ . Click the two intersection points; Desmos labels them $(0, -3)$ and $(2, 5)$ . Add their $y$ -values: $-3 + 5 = 2$ .

Example 2

Which of the following systems of equations has exactly one solution?

A) $y = x + 5$ and $y = x^2 + 3x + 5$

B) $y = x + 5$ and $y = x^2 + 3x + 6$

C) $y = x + 5$ and $y = x^2 + 3x + 7$

D) $y = x + 4$ and $y = x^2 + 3x + 6$

The strategy: for each, substitute the line into the parabola and check the discriminant of the resulting quadratic.

Each system reduces to $x + k = x^2 + 3x + c$ , i.e. $x^2 + 2x + (c - k) = 0$ . With $A = 1$ , $B = 2$ , $C = c - k$ , the discriminant is:

\Delta = B^2 - 4AC = 4 - 4(c - k)

Option	$k$	$c$	$c - k$	Quadratic	$\Delta = 4 - 4(c-k)$	Solutions
A	5	5	0	$x^2 + 2x = 0$	$4 - 4(0) = 4 > 0$	2
B	5	6	1	$x^2 + 2x + 1 = 0$	$4 - 4(1) = 0$	1 ✓
C	5	7	2	$x^2 + 2x + 2 = 0$	$4 - 4(2) = -4 < 0$	0
D	4	6	2	$x^2 + 2x + 2 = 0$	$4 - 4(2) = -4 < 0$	0

Option B detail: Setting $x + 5 = x^2 + 3x + 6$ gives $x^2 + 2x + 1 = 0$ . The discriminant is $\Delta = 2^2 - 4(1)(1) = 4 - 4 = 0$ , confirming exactly one solution. Factoring: $(x + 1)^2 = 0 \Rightarrow x = -1$ , so $y = -1 + 5 = 4$ . Verify in parabola: $(-1)^2 + 3(-1) + 6 = 1 - 3 + 6 = 4$ ✓. The line is tangent to the parabola at $(-1, 4)$ .

Why the distractors fail:

A) $\Delta = 4 > 0$ → two intersections, not one.
C) $\Delta = -4 < 0$ → no real intersections. A student who confuses " $\Delta = 0$ " with "no solution" might choose this.
D) Same quadratic as C, same error.

Answer: B)

Desmos check

Desmos check: Enter all four systems one at a time. The graph for option B shows the line touching the parabola at exactly one point; the others show zero or two.

Common Mistakes & Traps

Forgetting to rearrange before applying the discriminant. The discriminant applies to the quadratic after full substitution. Applying it to the parabola's original coefficients ignores the line.
Sign errors when collecting terms. Moving $mx + b$ across the equals sign changes every sign. Double-check the sign of the $x$ -coefficient in your combined quadratic.
Stopping at $x$ -values. The question often asks for $y$ -coordinates, a sum, or a product. Always back-substitute and re-read the question.
Confusing $\Delta = 0$ with $\Delta < 0$ . $\Delta = 0$ means one solution (tangency). $\Delta < 0$ means zero real solutions. These are frequently swapped in SAT distractors.
Substituting the parabola into itself. Make sure you substitute the linear expression for $y$ into the nonlinear equation, not the other way around.
Using Vieta's shortcut on the wrong quadratic. Vieta's $x_1 + x_2 = -B/A$ applies to the quadratic you get after substitution, not to the original parabola's equation.

Practice Questions

Question 1

How many solutions does the system $y = 3x - 2$ and $y = x^2 - x + 2$ have?

A) 0 B) 1 C) 2 D) Infinitely many

Show answer

Answer: B) 1

Substitute: $3x - 2 = x^2 - x + 2$

x^2 - 4x + 4 = 0 \implies (x - 2)^2 = 0

$\Delta = 0 \Rightarrow$ exactly one solution. At $x = 2$ : $y = 3(2) - 2 = 4$ . Check in parabola: $4 - 2 + 2 = 4$ ✓.

Why others fail:

A) 0: Confuses $\Delta = 0$ with no solutions.
C) 2: Student sees " $(x-2)^2 = 0$ " and thinks two different roots at $\pm 2$ .
D) Infinitely many: Misidentifies the system as having overlapping graphs.

Question 2 (Student-produced response)

The system of equations $y = x - 1$ and $y = x^2 - 4x + 3$ has two solutions. What is the sum of the $x$ -coordinates of the solutions?

Show answer

Answer: 5

Substitute: $x - 1 = x^2 - 4x + 3$

x^2 - 5x + 4 = 0 \implies (x - 4)(x - 1) = 0

$x = 4$ or $x = 1$ . Sum $= 4 + 1 = \mathbf{5}$ .

Verify: $(4,\ 3)$ : line gives $3$ ✓, parabola gives $16 - 16 + 3 = 3$ ✓. $(1,\ 0)$ : line gives $0$ ✓, parabola gives $1 - 4 + 3 = 0$ ✓.

Desmos check

Desmos check: Graph both equations; click the two intersections. The $x$ -labels are $1$ and $4$ . Their sum is $5$ .

Vieta's shortcut: From $x^2 - 5x + 4 = 0$ , sum of roots $= -(-5)/1 = 5$ .

Question 3

For what value of $k$ does the line $y = x + k$ intersect the parabola $y = x^2 - 2x + 5$ at exactly one point?

A) $-\dfrac{11}{4}$

B) $\dfrac{11}{4}$

C) $11$

D) $5$

Show answer

Answer: B) $\dfrac{11}{4}$

Substitute: $x + k = x^2 - 2x + 5 \implies x^2 - 3x + (5 - k) = 0$

For one intersection, set $\Delta = 0$ :

\Delta = (-3)^2 - 4(1)(5 - k) = 9 - 20 + 4k = 4k - 11 = 0 \implies k = \frac{11}{4}

Why others fail:

A) $-\frac{11}{4}$ : Student writes $x^2 - 3x + (5 + k) = 0$ (adds $k$ instead of subtracting), getting $-11 - 4k = 0 \Rightarrow k = -\frac{11}{4}$ .
C) $11$ : Student correctly forms $4k - 11 = 0$ but writes $k = 11$ , forgetting to divide by $4$ .
D) $5$ : Student sets $k$ equal to the constant term of the parabola rather than using the discriminant.

Question 4

Which of the following lists the solutions of the system $y = -x + 5$ and $y = x^2 - 4x + 5$ ?

A) $(0,\ 5)$ and $(3,\ 2)$

B) $(0,\ 5)$ and $(-3,\ 8)$

C) $(5,\ 0)$ and $(2,\ 3)$

D) $(0,\ 5)$ only

Show answer

Answer: A) $(0,\ 5)$ and $(3,\ 2)$

Substitute: $-x + 5 = x^2 - 4x + 5$

x^2 - 3x = 0 \implies x(x - 3) = 0 \implies x = 0 \text{ or } x = 3

$x = 0$ : $y = 5$ . Parabola: $0 - 0 + 5 = 5$ ✓
$x = 3$ : $y = 2$ . Parabola: $9 - 12 + 5 = 2$ ✓

Why others fail:

B) Uses $x = -3$ to compute $y$ : $-(-3) + 5 = 8$ , confusing the sign of the second root.
C) Swaps the $x$ - and $y$ -coordinates of each solution.
D) Factors $x(x-3) = 0$ but records only $x = 0$ , missing the second solution.

Question 5 (Student-produced response)

For the system $y = 5x - 3$ and $y = x^2 + 3x - 3$ , the two solutions are $(0,\ p)$ and $(2,\ q)$ . What is the value of $p + q$ ?

Show answer

Answer: 4

Substitute: $5x - 3 = x^2 + 3x - 3$

x^2 - 2x = 0 \implies x(x - 2) = 0 \implies x = 0 \text{ or } x = 2

$x = 0$ : $y = 5(0) - 3 = -3$ , so $p = -3$ . Check: $0 + 0 - 3 = -3$ ✓
$x = 2$ : $y = 5(2) - 3 = 7$ , so $q = 7$ . Check: $4 + 6 - 3 = 7$ ✓

p + q = -3 + 7 = \mathbf{4}

Connections

Solving Quadratic Equations (prerequisite): factoring and the quadratic formula are the core tools once substitution produces the quadratic. Weakness there will surface here.
Solving Systems: Substitution and Elimination (prerequisite): this note extends that method — the substitution step is identical; the new element is that one equation is nonlinear.
The Discriminant and Number of Solutions (sibling note): covers $\Delta$ in depth for single quadratics; here we apply $\Delta$ to the quadratic formed by merging the two equations.
Absolute Value, Radical, and Rational Equations (sibling note): similar substitution logic applies to other nonlinear types.
Solving Formulas for a Variable (sibling note): symbolic manipulation skills that keep the algebra clean during the collection step.
On test day, this skill often combines with interpreting a graph: after solving algebraically, you may be asked to identify which graph correctly shows the number of intersections — your discriminant tells you immediately what to look for.

Introduction

Core Concept

What "solving the system" means geometrically

The substitution strategy

Key Formulas & Rules

Worked Examples

Example 1

Example 2

Common Mistakes & Traps

Practice Questions

Connections

Figures

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