Digital SAT Math · Advanced Math

Nonlinear systems

A line and a parabola share zero, one, or two points — and after one substitution that count is just the discriminant of a quadratic you already know how to read. The expensive mistake on this skill is not the algebra. It is finishing the $x$ -values and bubbling them when the stem asked for a point, a $y$ , or a count you could have answered without solving at all.

On the test


Domain	Advanced Math (score report)
What it looks like	Two equations in $x$ and $y$ , at least one nonlinear — usually a line with a parabola, sometimes a line with a circle
Often asked	“What is a solution of the system…?”, “How many real solutions…?”, “For what value of $k$ does the system have exactly one solution?”, “What is the $y$ -coordinate of…?”
Format	Multiple choice and student-produced response
Calculator	Desmos graphs both equations and marks the crossings; for “how many” and for ugly coefficients it is often fastest — see the box below

Recognition cues: two equations stacked, one carrying an $x^2$ or an $x^2 + y^2$ ; the phrases exactly one solution, no real solutions, ordered pair $(x, y)$ ; a constant $k$ controlling whether the graphs touch; and any stem that hands you a coordinate plane with a curve and a line.

What this page owns

The College Board files “nonlinear equations in one variable” and “systems of equations in two variables” under one heading. We split them. Here there are two equations and two unknowns, and the answer is an intersection point, a coordinate of one, a count of them, or the constant that forces a count. One variable and one equation lives on nonlinear equations. Two linear equations live on systems of linear equations. A parabola whose features you read without ever writing a second equation lives on quadratic functions. The substitution step is the same idea as linear systems; the quadratic that falls out is the same object as the one-variable page — what changes is that you stop, or you don’t, at the discriminant.

Pattern recognition

Five shapes cover nearly every item:

Both solved for $y$ — $y = \ldots$ twice. Set the right sides equal.
Linear equation easy to rearrange — isolate $x$ or $y$ from the line, substitute into the curve.
Count only — “how many solutions” or “exactly one / none / two”. Discriminant after sub; do not solve.
Parameter $k$ — a letter in one equation, and the stem fixes the count. Set the discriminant to match that count and solve for the letter.
Graph handed to you — a parabola and a line already drawn; read crossings, or recognise a tangency.

Method

Name the ask first. Underline how many, exactly one, $k$ , $x$ , $y$ , or the ordered pair. The ask decides whether you stop at the discriminant.
Substitute. Prefer the linear equation as the source of the replacement — less arithmetic, and it is the safer place to recover $y$ later.
Standard form. One side zero before you name $a$ , $b$ , and $c$ . $x^2 = 2x + 3$ is not ready; $x^2 - 2x - 3 = 0$ is.
Branch. Count → discriminant. Coordinates → factor or formula, then back-sub $y$ from the line.
Answer the combination that was named. The value of $x$ alone is almost always a listed choice. So is the other intersection’s coordinate, and so is “zero” when the line is tangent.

What you see after sub	Tool	Stop when
stem says “how many” / “exactly one” / “no real”	discriminant $b^2 - 4ac$	you have the sign (or the zero)
stem hides a letter $k$ and fixes the count	set disc $= 0$ (one) or write an inequality (two / none)	$k$ is found — never solve for $x$
stem wants $x$ , $y$ , or $(x, y)$	solve the quadratic, back-sub into the line	both coordinates match the ask
graph already drawn	count crossings; tangency = one	you have read the picture

Worked example 1 — substitute, finish the point, answer $y$

Stem.

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

If $(x, y)$ is a solution of the system and $x > 0$ , what is the value of $y$ ?

Step 1 — set the expressions equal. Both are already solved for $y$ :

x^2 = 2x + 3 \qquad \Rightarrow \qquad x^2 - 2x - 3 = 0

Step 2 — factor. $(x - 3)(x + 1) = 0$ , so $x = 3$ or $x = -1$ . The condition $x > 0$ keeps $x = 3$ .

Step 3 — back-sub into the line. $y = 2(3) + 3 = 9$ . The point is $(3, 9)$ .

Check. On the parabola, $3^2 = 9$ ✓. On the line, $2\cdot 3 + 3 = 9$ ✓. The discarded root gives $(-1, 1)$ , which also solves the system but fails $x > 0$ . Answer: 9.

Trap watch. Bubbling $3$ — the positive $x$ you just found — is the single most common loss on this shape. The stem named $y$ .

Worked example 2 — stop at the discriminant

Stem.

y = x^2 + 4 \qquad y = 2x - 1

How many real solutions does the system have?

Step 1 — substitute.

x^2 + 4 = 2x - 1 \qquad \Rightarrow \qquad x^2 - 2x + 5 = 0

Step 2 — do not solve. The stem asked for a count:

b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16 < 0

Negative discriminant means no real $x$ , so no real ordered pairs.

Check. Completing the square: $x^2 - 2x + 5 = (x - 1)^2 + 4 \ge 4 > 0$ , so the left side after moving terms never hits zero — same conclusion without the formula. Graphically the line sits entirely below the upward parabola. Answer: zero real solutions.

Trap watch. Running the quadratic formula anyway produces complex roots and wastes time; “one solution” is the reading that confuses a never-touching pair with a tangency.

Worked example 3 — a constant that forces a tangency

Stem.

y = x^2 \qquad y = 2x + k

$k$ is a constant. For what value of $k$ does the system have exactly one real solution?

Step 1 — substitute and arrange.

x^2 = 2x + k \qquad \Rightarrow \qquad x^2 - 2x - k = 0

Step 2 — exactly one solution means the discriminant is zero.

(-2)^2 - 4(1)(-k) = 0 \qquad \Rightarrow \qquad 4 + 4k = 0 \qquad \Rightarrow \qquad k = -1

Check. With $k = -1$ the quadratic is $x^2 - 2x + 1 = (x - 1)^2 = 0$ , so $x = 1$ only, and $y = 2(1) - 1 = 1$ . The line $y = 2x - 1$ is tangent to $y = x^2$ at $(1, 1)$ . For $k = 0$ , disc $= 4 > 0$ (two solutions); for $k = -2$ , disc $= -4 < 0$ (none). Answer: $k = -1$ .

Trap watch. Dropping the sign when expanding $4ac$ with $c = -k$ turns $4 + 4k$ into $4 - 4k$ and answers $k = 1$ , whose system has two solutions. Treating “exactly one” as “no solution” is the geometric error from the callout above.

Practice

Answer before opening the explanation. Two items are student-produced response, matching the grid-in share of the live digital sets on this band; two hand you a line and a parabola already drawn. Every wrong choice below is one named misconception — only $x$ reported, $y$ recovered from the wrong equation, a discriminant sign flipped, a tangency counted as none. When you miss one, log the misconception, not the item number.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

What are all values of x that satisfy the system y = x + 3 and y = x² + 1?

Show the answer Choice C

Why it is right

Both equations give y, so set the right sides equal: x² + 1 = x + 3. Move everything to one side to get x² - x - 2 = 0. Factor as (x - 2)(x + 1) = 0, so x = 2 or x = -1. Checking: at x = 2, both equations give y = 5; at x = -1, both give y = 2. The system has two solutions, and the x-coordinates are -1 and 2.

Why each other choice fails

Choice A: Flips the signs of the true roots. Setting (x - 2)(x + 1) = 0 produces +2 and -1, not -2 and +1. At x = -2 the two y-expressions are 1 and 5, which do not match.
Choice B: Keeps the positive root 2 but replaces -1 with +1. At x = 1, the line gives y = 4 while the parabola gives y = 2, so x = 1 is not a solution.
Choice D: Drops the root x = 2 after factoring. Both factors produce a valid intersection: (2, 5) sits on both graphs just as (-1, 2) does.

Question 2 Standard

One solution of the system y = 2x − 1 and y = x² − 4x + 7 is the ordered pair (2, y). What is the value of y?

Show the answer Choice B

Why it is right

Substitute x = 2 into the linear equation: y = 2(2) − 1 = 3. The point is (2, 3). Confirming the long way, set the expressions equal: x² − 4x + 7 = 2x − 1 gives x² − 6x + 8 = 0, which factors as (x − 2)(x − 4) = 0. The roots are x = 2 and x = 4; at x = 2 the line (or the parabola) returns y = 3, so (2, 3) solves the system.

Why each other choice fails

Choice A: Reports the given x-coordinate instead of finishing the ordered pair. The stem already supplied x = 2 and asked for the matching y.
Choice C: Is the y-coordinate of the other intersection (4, 7), or the constant term of the parabola. At x = 2 the parabola is 4 − 8 + 7 = 3, not 7.
Choice D: Comes from using y = 2x + 1 (sign error on the constant) or from evaluating 2(2) + 1. The linear equation is y = 2x − 1, so the constant is subtracted, not added.

Question 3 Standard

How many real solutions does the system y = x + 5 and y = x² − 3x + 1 have?

Show the answer Choice A

Why it is right

Set the right sides equal: x² − 3x + 1 = x + 5, so x² − 4x − 4 = 0. The stem asks for a count, not the roots, so compute the discriminant: (−4)² − 4(1)(−4) = 16 + 16 = 32. A positive discriminant means two distinct real values of x, and each produces a y from the line, so the system has two real solutions. (Solving is unnecessary here; the quadratic formula would only recover the same two x-values.)

Why each other choice fails

Choice B: Is the reading of a zero discriminant (tangency). Here Δ = 32 ≠ 0, so the line crosses the parabola twice rather than grazing it once.
Choice C: Is the reading of a negative discriminant. Adding 16 and 16 gives +32, not a negative value; the graphs do meet.
Choice D: Would require the two equations to describe the same curve for every x. A non-degenerate parabola and a line never coincide on an infinite set of points.

Question 4 Standard Student-produced response

The system y = 6x − 9 and y = x² has exactly one real solution. What is the x-coordinate of that solution?

Your answer

Show the answer 3

Why it is right

Set the expressions equal: x² = 6x − 9, so x² − 6x + 9 = 0. The left side is a perfect square: (x − 3)² = 0, which has the double root x = 3. That single x-value is the tangency: the line meets the parabola at exactly one point. Back-substituting into the line gives y = 6(3) − 9 = 9, so the solution is (3, 9), and the x-coordinate the stem asked for is 3. The discriminant of x² − 6x + 9 is 36 − 36 = 0, confirming exactly one real solution.

Answers students type instead

6: Reports the slope of the line, which is a coefficient in the system but not a root of the quadratic after substitution.
9: Reports the y-coordinate of the tangency point, or the constant on the right side of the original line equation. The stem asked for the x-coordinate.
-3: Flips the sign of the double root. Expanding (x + 3)² gives x² + 6x + 9, which is not the arranged equation x² − 6x + 9.

Question 5 Standard

The solutions of the system y = 8 − 2x and y = x² are the ordered pairs (−4, 16) and (2, y). What is the value of y?

Show the answer Choice D

Why it is right

At the solution with x = 2, the linear equation gives y = 8 − 2(2) = 4. Checking the parabola: 2² = 4, so both equations agree at (2, 4). The full solve confirms the pair of roots: x² = 8 − 2x rearranges to x² + 2x − 8 = 0, which factors as (x + 4)(x − 2) = 0, so the intersections are (−4, 16) and (2, 4).

Why each other choice fails

Choice A: Reports the x-coordinate of the same intersection instead of its y-coordinate. The stem already named x = 2 and asked for y.
Choice B: Reports the x-coordinate of the other intersection. That other point is (−4, 16); its x is not a y-value of either solution.
Choice C: Is the y-coordinate of the other intersection (−4, 16). At x = 2 the line (and the parabola) give 4, not 16.

Question 6 Harder

The graphs of y = x² − 2x − 3 and y = x + 1 are shown in the xy-plane. Which ordered pair is a solution of the system?

Figure 1. The graphs of y = x² − 2x − 3 and y = x + 1, with both intersection points marked.

Show the answer Choice B

Why it is right

A solution is a point that lies on both graphs — one of the two marked crossings. The right-hand crossing sits at x = 4 and y = 5, so the ordered pair is (4, 5). Algebra confirms it: set x² − 2x − 3 = x + 1 to get x² − 3x − 4 = 0, which factors as (x − 4)(x + 1) = 0. At x = 4 the line gives y = 5, matching the marked point. (The other solution is (−1, 0), which is not among the choices.)

Why each other choice fails

Choice A: Swaps the coordinates of the true intersection (4, 5). An ordered pair is (horizontal, vertical), so 4 across and 5 up is (4, 5), not (5, 4). At x = 5 the parabola is 12 and the line is 6 — neither is 4.
Choice C: Uses the correct x-coordinate of the left intersection but the wrong y. At x = −1 both graphs give y = 0, not y = −2. The point (−1, −2) sits on neither graph.
Choice D: Is the y-intercept of the parabola alone (when x = 0, y = −3). The line at x = 0 gives y = 1, so (0, −3) is not an intersection.

Question 7 Harder

How many real solutions does the system y = x² + 4 and y = 2x − 1 have?

Show the answer Choice C

Why it is right

Substitute: x² + 4 = 2x − 1 rearranges to x² − 2x + 5 = 0. The discriminant is (−2)² − 4(1)(5) = 4 − 20 = −16, which is negative, so there are no real values of x and therefore no real ordered pairs that satisfy both equations. Completing the square gives the same picture: x² − 2x + 5 = (x − 1)² + 4, which is always at least 4 and never zero. The line lies entirely below the upward-opening parabola.

Why each other choice fails

Choice A: Is the count when the discriminant is positive. Here 4 − 20 is negative, not positive, so the graphs never meet.
Choice B: Is the count when the discriminant is zero (a tangency). A zero discriminant would require 4 − 20 = 0, which is false; the line misses the parabola rather than grazing it.
Choice D: Would require the two equations to be the same relation. A parabola y = x² + 4 and a non-matching line cannot share infinitely many points.

Question 8 Harder

The graphs of y = x² and y = 2x − 1 are shown in the xy-plane. How many real solutions does the system have?

Figure 2. The graphs of y = x² and y = 2x − 1, which touch at exactly one point.

Show the answer Choice A

Why it is right

The figure shows the line touching the parabola at a single marked point (1, 1) — a tangency. Algebra matches the picture: set x² = 2x − 1 to get x² − 2x + 1 = 0, which is (x − 1)² = 0. A repeated root is still exactly one value of x, and with y = 2(1) − 1 = 1 the system has exactly one real solution. The discriminant is 4 − 4 = 0, the algebraic form of the same fact.

Why each other choice fails

Choice B: Would be correct if the line cut through the parabola at two distinct points. Here the graphs share only the tangency point; the discriminant is zero, not positive.
Choice C: Is the trap that treats a graze as a miss. Touching at one point is one solution, not zero. Zero solutions require a negative discriminant, and here Δ = 0.
Choice D: Would require the parabola and the line to be the same graph. They share only the single point (1, 1).

Question 9 Harder

Which ordered pair is a solution of the system x² + y² = 25 and y = x − 1?

Show the answer Choice B

Why it is right

Substitute y = x − 1 into the circle: x² + (x − 1)² = 25 expands to 2x² − 2x + 1 = 25, so 2x² − 2x − 24 = 0, or x² − x − 12 = 0. Factoring gives (x − 4)(x + 3) = 0, so x = 4 or x = −3. Back-substitute into the line (not into the expanded quadratic): at x = 4, y = 4 − 1 = 3, giving the point (4, 3). Check on the circle: 16 + 9 = 25 ✓. The other solution is (−3, −4), which is not among the choices.

Why each other choice fails

Choice A: Sits on the circle (16 + 9 = 25) but not on the line: the line at x = 4 gives y = 3, not −3. The error is recovering y from the wrong equation or flipping the sign after back-substitution — a point must satisfy both equations, and the line is the cheap place to read y.
Choice C: Swaps the coordinates of the true solution (4, 3). The pair (3, 4) fails the line (4 ≠ 3 − 1) and the circle (9 + 16 = 25 holds, but the line does not).
Choice D: Mixes the x-coordinate of the other intersection with a positive y. At x = −3 the line gives y = −4, not +4. The point (−3, 4) fails both equations.

Question 10 Hardest Student-produced response

In the system y = kx + 1 and y = x² + 5, k is a constant. The system has exactly one real solution. What is the positive value of k?

Your answer

Show the answer 4

Why it is right

Set the expressions equal: x² + 5 = kx + 1, so x² − kx + 4 = 0. Exactly one real solution means the discriminant is zero: (−k)² − 4(1)(4) = 0, so k² − 16 = 0 and k = ±4. The stem asks for the positive value, so k = 4. Checking: with k = 4 the quadratic is x² − 4x + 4 = (x − 2)² = 0, a double root at x = 2, and y = 4(2) + 1 = 9. The line y = 4x + 1 is tangent to the parabola y = x² + 5 at (2, 9).

Answers students type instead

2: Is the x-coordinate of the tangency point when k = 4, or comes from dropping a factor of 4 in the discriminant (solving k² − 4 = 0). The stem asked for k, not for x.
16: Reports k² rather than k. The discriminant equation is k² − 16 = 0, so k² equals 16 and k itself is ±4.
-4: The other root of k² = 16. It also produces a tangency (the line with slope −4), but the stem restricted the answer to the positive value of k.

Question 11 Hardest

What is the sum of the y-coordinates of the solutions of the system y = 4x − 5 and y = x² + x − 5?

Show the answer Choice D

Why it is right

Set the expressions equal: x² + x − 5 = 4x − 5, so x² − 3x = 0, or x(x − 3) = 0. The solutions are x = 0 and x = 3. Back-substitute into the line: at x = 0, y = −5; at x = 3, y = 4(3) − 5 = 7. The two y-coordinates are −5 and 7, and their sum is 2. (As a shortcut, the sum of the y-values from y = 4x − 5 at the two roots is 4(0 + 3) − 5 − 5 = 12 − 10 = 2, matching.)

Why each other choice fails

Choice A: Is the sum of the x-coordinates (0 + 3 = 3). The stem asked for the sum of the y-coordinates, which is a different number.
Choice B: Is one of the x-roots, or the product of the x-roots. Neither is the sum of the y-values.
Choice C: Is the y-coordinate of only the first intersection (0, −5). The second intersection contributes y = 7, and −5 + 7 = 2.

Question 12 Hardest

In the system y = x² − 6x + k and y = 2x + 1, k is a constant. For what value of k does the system have exactly one real solution?

Show the answer Choice A

Why it is right

Substitute: x² − 6x + k = 2x + 1 rearranges to x² − 8x + (k − 1) = 0. Exactly one real solution means the discriminant is zero: (−8)² − 4(1)(k − 1) = 0, so 64 − 4k + 4 = 0, hence 68 = 4k and k = 17. Checking: with k = 17 the quadratic is x² − 8x + 16 = (x − 4)² = 0, a double root at x = 4, and y = 2(4) + 1 = 9. The graphs touch at exactly one point.

Why each other choice fails

Choice B: Comes from treating the constant term as k rather than k − 1 — that is, setting 64 − 4k = 0 and getting k = 16. After moving the line's constant, the constant term of the quadratic is k − 1, so the 4 in 4ac carries an extra −1 that shifts the answer by 1.
Choice C: Is the x-coordinate of the tangency point (or half of 8 from −b/2a). The stem asked for the constant k, not for the x-value where the graphs touch.
Choice D: Is the constant term of the line y = 2x + 1, or the result of setting k − 1 = 0. That value of k makes the constant term vanish but leaves a positive discriminant (64), so two solutions.

Common mistakes

Reporting $x$ when the stem asked for $y$ or for the point — the quadratic produces $x$ -values; the ordered pair still needs the linear equation. Expect the bare $x$ to sit among the choices.
Back-substituting into the quadratic you just built — it is true by construction and checks nothing. Recover $y$ from the untouched line.
Reading the discriminant off an unarranged equation — $x^2 = 2x + 3$ is not yet $a$ , $b$ , $c$ . Move everything to one side first.
Calling a tangency “no solution” — $\Delta = 0$ is exactly one ordered pair, the graze point. “None” is $\Delta < 0$ .
Solving in full when only a count was asked — “how many real solutions” is one subtraction after sub. The formula is extra work that invites arithmetic slips.
Losing a root of the quadratic — after sub you still have a two-root equation; both $x$ -values are intersections unless a stem condition (positive $x$ , a quadrant) kills one.
Sign error in $4ac$ when $c$ carries a letter — for $x^2 - 2x - k = 0$ , $c = -k$ , so $4ac = -4k$ . Writing $4 + 4k = 0$ versus $4 - 4k = 0$ is the whole item.
Swapping coordinates of the ordered pair — $(3, 9)$ is not $(9, 3)$ . On a graph, horizontal then vertical.
Using the nonlinear equation for a cheap $y$ when the line is sitting right there — both are valid, but $y = 2x + 3$ is one multiplication; $y = x^2$ is a square that invites a slip on a negative root.
Treating a circle-and-line system as if it were linear — after sub you still get a quadratic; the same discriminant rules apply, and points must satisfy $x^2 + y^2 = r^2$ .
Answering “infinitely many” for a line on a parabola — a non-degenerate parabola is not a line, so the graphs never coincide for every $x$ . Degenerate counts on this skill are zero, one, or two.
Forgetting the domain on a stem that restricts the answer — “positive $x$ ”, “the solution in the first quadrant”, “the greater $y$ ”. Both roots may be algebraically fine and only one is the answer.

FAQ

Is elimination ever better than substitution on a nonlinear system? Almost never on this test. Elimination needs matching degrees; a line and a parabola do not cancel cleanly. Isolate from the line, substitute, done.

Can a line and a parabola have more than two intersections? Not if the nonlinear piece is genuinely quadratic (degree 2). After sub you hold a degree-2 equation, which has at most two real roots. Higher-degree curves can do more; they are rare on the Digital SAT.

What if neither equation is solved for a variable? Pick the linear one and solve for $x$ or $y$ yourself — one line of algebra — then substitute. Do not expand a product of two messy expressions if a linear isolation is available.

Desmos or discriminant for “how many solutions”? Both work. Discriminant is faster when the coefficients are clean integers and you already substituted. Desmos is faster when the equations are ugly or already typed. On a parameter item, algebra wins for the exact value.

The system is a circle and a line. Same method? Yes: isolate from the line, substitute into $x^2 + y^2 = r^2$ , collect to a quadratic, then count or solve. Back-sub $y$ from the line. Check that both coordinates satisfy the circle equation.

Grid-in format? Enter the exact value the stem names — often a single coordinate, a sum, or a value of $k$ . Fractions are accepted as fractions; the most common grid-in loss is entering $x$ when the item asked for $y$ .

How many of these show up on a test? Nonlinear equations and systems together are a large piece of Advanced Math. Free guides put the pair at several questions across the two modules; the one-variable half is drilled on its own page, and this page owns the two-variable half.

Editorial team · last reviewed 2026-08-12

Nonlinear systems

On the test

What this page owns

Pattern recognition

Method

Worked example 1 — substitute, finish the point, answer yyy

Worked example 2 — stop at the discriminant

Worked example 3 — a constant that forces a tangency

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

Why it is right

Why each other choice fails

Question 3 Standard

Why it is right

Why each other choice fails

Question 4 Standard Student-produced response

Why it is right

Answers students type instead

Question 5 Standard

Why it is right

Why each other choice fails

Question 6 Harder

Why it is right

Why each other choice fails

Question 7 Harder

Why it is right

Why each other choice fails

Question 8 Harder

Why it is right

Why each other choice fails

Question 9 Harder

Why it is right

Why each other choice fails

Question 10 Hardest Student-produced response

Why it is right

Answers students type instead

Question 11 Hardest

Why it is right

Why each other choice fails

Question 12 Hardest

Why it is right

Why each other choice fails

Common mistakes

FAQ

Related skills

Worked example 1 — substitute, finish the point, answer $y$