Algebra 2 Quiz: Solving Linear Quadratic Systems

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Solving Linear Quadratic SystemsQuestion 1 of 20

Find the intersection point(s) of the line and the circle by solving the system.

\begin{cases} x^2 + y^2 = 25 \\ y = -3x \end{cases}

What are all solutions $(x,y)$ ?

\left(\frac{5}{10},-\frac{15}{10}\right)

and

\left(-\frac{5}{10},\frac{15}{10}\right)

(1,-3)

and

(-1,3)

\left(\frac{5}{\sqrt{10}},-\frac{15}{\sqrt{10}}\right)

and

\left(-\frac{5}{\sqrt{10}},\frac{15}{\sqrt{10}}\right)

\left(\frac{5}{\sqrt{10}},\frac{15}{\sqrt{10}}\right)

and

\left(-\frac{5}{\sqrt{10}},-\frac{15}{\sqrt{10}}\right)

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Algebra 2 Quiz

Algebra 2 Quiz: Solving Linear Quadratic Systems

Practice Solving Linear Quadratic Systems in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Linear Quadratic Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Find the intersection point(s) of the line and the circle by solving the system.

\begin{cases} x^2 + y^2 = 25 \\ y = -3x \end{cases}

What are all solutions $(x,y)$ ?

$\left(\frac{5}{10},-\frac{15}{10}\right)$ and $\left(-\frac{5}{10},\frac{15}{10}\right)$
$(1,-3)$ and $(-1,3)$
$\left(\frac{5}{\sqrt{10}},-\frac{15}{\sqrt{10}}\right)$ and $\left(-\frac{5}{\sqrt{10}},\frac{15}{\sqrt{10}}\right)$ (correct answer)
$\left(\frac{5}{\sqrt{10}},\frac{15}{\sqrt{10}}\right)$ and $\left(-\frac{5}{\sqrt{10}},-\frac{15}{\sqrt{10}}\right)$

Explanation: This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (circle) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve like a circle: the solutions are the intersection points, which can be 0, 1, or 2 depending on whether the line misses, touches, or crosses the circle. For x² + y² = 25 and y = -3x, substitute: x² + (-3x)² = 25 becomes 10x² = 25, so x² = 2.5, x = ±√(5/2) = ±5/√10 after simplifying; then y = -3x gives (5/√10, -15/√10) and (-5/√10, 15/√10), both satisfying the equations. Choice A correctly finds both points with proper substitution and radical simplification. Choice C might appeal if you incorrectly simplify √(25/10) to 1 and -1, but remember to handle the square root accurately—x² = 25/10 means x = ±5/√10! Always substitute the linear into the quadratic, solve for the variable, back-substitute, and check the discriminant for the number of solutions. Keep up the excellent work—this approach will help you tackle any linear-circle system confidently!

Question 2

Solve the system. The solution(s) are the intersection point(s) of the line and the parabola.

\begin{cases} y = -x + 2 \\ y = x^2 \end{cases}

$(1,1)$ and $(-2,4)$ (correct answer)
$(1,1)$ and $(2,4)$
$(0,2)$ and $(2,0)$
$(-1,1)$ only

Explanation: This question tests your ability to solve linear-quadratic systems to find intersection points of a line and parabola. Such systems can yield 0, 1, or 2 solutions depending on intersections. For

y = -x + 2

and

y = x^2

, set equal:

x^2 = -x + 2

x^2 + x - 2 = 0

; discriminant

1 + 8 = 9

x = \frac{-1 \pm 3}{2} = 1

-2

;

y = 1

and

4

, giving

(1,1)

and

(-2,4)

, both checked. Choice A correctly lists the points. Choice B might come from sign errors in quadratic, but track coefficients carefully! Use the method: substitute, form quadratic, solve with formula or factoring, back-substitute, verify. Keep going—you're building expertise step by step!

Question 3

Find all intersection point(s) of the line and the circle (solve the system). You may solve algebraically and confirm graphically.

\begin{cases} x^2 + y^2 = 25 \\ y = -3x \end{cases}

What are the solution(s) $(x,y)$ ?

$\left(\frac{5}{\sqrt{10}},-\frac{15}{\sqrt{10}}\right)$ and $\left(-\frac{5}{\sqrt{10}},\frac{15}{\sqrt{10}}\right)$ (correct answer)
$\left(\frac{5}{\sqrt{10}},\frac{15}{\sqrt{10}}\right)$ and $\left(-\frac{5}{\sqrt{10}},-\frac{15}{\sqrt{10}}\right)$
$\left(\frac{5}{10},-\frac{15}{10}\right)$ and $\left(-\frac{5}{10},\frac{15}{10}\right)$
$(0,5)$ and $(0,-5)$

x^2 + y^2 = 25

, yielding

x^2 + 9x^2 = 25

, so

10x^2 = 25

x^2 = 2.5

x = \pm \sqrt{\frac{5}{2}} = \pm \frac{5}{\sqrt{10}}

; then y = -3x gives

(\frac{5}{\sqrt{10}}, -\frac{15}{\sqrt{10}})

and

(-\frac{5}{\sqrt{10}}, \frac{15}{\sqrt{10}})

. Choice A correctly finds both intersection points with accurate substitution and rationalized forms. A common mistake is forgetting to rationalize or mixing signs in y-values, but double-checking with the circle equation ensures correctness. Keep practicing: visualize the line through the origin with slope -3 crossing the circle centered at origin with radius 5—great job verifying graphically too!

Question 4

Verify whether the point $(2,3)$ is a solution to the linear-quadratic system (a solution must satisfy both equations).

\begin{cases} y = x^2 - 1 \\ y = 3x - 3 \end{cases}

Which statement is correct?

Yes, because it satisfies both equations. (correct answer)
No, because it satisfies neither equation.
No, because it satisfies only $y=x^2-1$ .
No, because it satisfies only $y=3x-3$ .

Explanation: This question tests your ability to verify if a given point satisfies both equations in a linear-quadratic system, meaning it's an intersection point. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution, but for verification, plug the point into both equations and check if they hold true. For the point

(2, 3)

y = x^2 - 1

and

y = 3x - 3

: first equation

2^2 - 1 = 3

, second

3(2) - 3 = 3

, so yes, it satisfies both. Choice A correctly confirms it works for both. You're awesome—verification is a key step to catch errors after solving!

Question 5

Solve the linear-quadratic system algebraically (by substitution). The solutions correspond to the intersection point(s) of the line and the parabola.

\begin{cases} y = x + 1 \\ y = x^2 - 2x + 1 \end{cases}

What are all solutions $(x,y)$ to the system?

$(0,1)$ only
$(0,1)$ and $(3,4)$ (correct answer)
No real solution
$(1,2)$ and $(2,3)$

y = x + 1

and

y = x^2 - 2x + 1

, substitute the linear into the quadratic:

x + 1 = x^2 - 2x + 1

, rearrange to

x^2 - 3x = 0

, factor as

x(x - 3) = 0

, so

x = 0

x = 3

; then

y = 1

and

y = 4

, giving points

(0, 1)

and

(3, 4)

, which both satisfy the original equations. Choice C correctly identifies both intersection points through accurate substitution and back-substitution. Choice A might tempt if you mistakenly discard

x = 3

, but remember quadratics often yield two solutions—check both! To master linear-quadratic systems, always substitute the linear expression into the quadratic, solve the resulting quadratic equation, back-substitute to find y-values, and verify each point in both equations. You're doing great—practicing this method will make solving these systems second nature!

Question 6

Verify whether the point $(2,3)$ is a solution to the linear-quadratic system (a solution must satisfy both equations).

\begin{cases} y = x^2 - 1 \\ y = 3x - 3 \end{cases}

Which statement is correct?

No, because it satisfies neither equation.
Yes, because it satisfies both equations. (correct answer)
No, because it satisfies only $y=3x-3$ .
No, because it satisfies only $y=x^2-1$ .

(2, 3)

y = x^2 - 1

and

y = 3x - 3

: first equation

2^2 - 1 = 3

, second

3(2) - 3 = 3

, so yes, it satisfies both. Choice A correctly confirms it works for both. You're awesome—verification is a key step to catch errors after solving!

Question 7

Solve the system algebraically. The solution(s) are the intersection point(s) of the line and the parabola.

\begin{cases} y = -x + 1 \\ y = x^2 + 2x - 3 \end{cases}

How many real solutions does the system have?

0 real solutions
1 real solution
2 real solutions (correct answer)
Infinitely many solutions

Explanation: This question tests your ability to determine the number of real solutions in a linear-quadratic system, corresponding to intersection points of a line and parabola. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: solve the linear equation for y (often already in y = mx + b form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system y = -x + 1 and y = x² + 2x - 3, substitute: -x + 1 = x² + 2x - 3, rearrange to x² + 3x - 4 = 0, discriminant 9 + 16 = 25 > 0, so two real solutions (x = 1 and x = -4). Choice C correctly identifies there are 2 real solutions based on the positive discriminant. Excellent—using the discriminant is a quick way to preview the number without full solving!

Question 8

The system $\begin{cases} x^2 + y^2 = r^2 \\ y = x + c \end{cases}$ represents a circle and a line. If the system has no real solutions, which of the following must be true about the relationship between $r$ and $c$ ?

$|c| > r\sqrt{2}$ and $r > 0$ (correct answer)
$|c| < r\sqrt{2}$ and $r > 0$
$|c| > \frac{r}{\sqrt{2}}$ and $r > 0$
$|c| = r\sqrt{2}$ and $r > 0$

Explanation: For no real solutions, the line y = x + c must not intersect the circle x² + y² = r². Substituting: x² + (x + c)² = r², which gives x² + x² + 2cx + c² = r², or 2x² + 2cx + (c² - r²) = 0. For no real solutions, the discriminant must be negative: (2c)² - 4(2)(c² - r²) < 0. This gives 4c² - 8c² + 8r² < 0, so -4c² + 8r² < 0, which means 4c² > 8r², or c² > 2r². Taking square roots: |c| > r√2. We also need r > 0 for a valid circle. Choice B incorrectly uses < instead of >. Choice C uses r/√2 instead of r√2. Choice D represents the boundary case where the line is tangent to the circle.

Question 9

A system consists of the equations $y = x^2 - 4x + 3$ and $y = 2x - 5$ . If the solutions to this system are plotted on a coordinate plane, what is the sum of the x-coordinates of all intersection points?

2
4
6 (correct answer)
8

Explanation: Setting the equations equal: x² - 4x + 3 = 2x - 5. Rearranging: x² - 6x + 8 = 0. Factoring: (x - 2)(x - 4) = 0, so x = 2 and x = 4. The sum is 2 + 4 = 6. Choice A gives only one x-coordinate. Choice B is the coefficient error from x² - 4x + 8 = 0. Choice D results from incorrectly rearranging to x² - 6x + 16 = 0.

Question 10

A parabola and a line intersect at points $A$ and $B$ . If the parabola is $y = x^2 + 2x - 3$ and the midpoint of segment $AB$ is $(1, 2)$ , what is the equation of the line?

$y = 4x - 2$ (correct answer)
$y = 3x - 1$
$y = 2x + 0$
$y = 5x - 3$

Explanation: Let the line be y = mx + b, and let the intersection points be (x₁, y₁) and (x₂, y₂). Since the midpoint is (1, 2), we have (x₁ + x₂)/2 = 1, so x₁ + x₂ = 2. Setting x² + 2x - 3 = mx + b gives x² + (2 - m)x + (-3 - b) = 0. By Vieta's formulas, x₁ + x₂ = -(2 - m) = m - 2. Since x₁ + x₂ = 2, we have m - 2 = 2, so m = 4. For the y-coordinate of the midpoint: (y₁ + y₂)/2 = 2, so y₁ + y₂ = 4. Since both points lie on the line y = 4x + b, we have y₁ = 4x₁ + b and y₂ = 4x₂ + b. Thus y₁ + y₂ = 4(x₁ + x₂) + 2b = 4(2) + 2b = 8 + 2b. Setting this equal to 4: 8 + 2b = 4, so 2b = -4, giving b = -2. Therefore, the line is y = 4x - 2. Choice B gives slope 3, not 4. Choice C gives slope 2, not 4. Choice D has the correct slope but wrong y-intercept.

Question 11

Solve the linear-quadratic system algebraically (by substitution). The solution(s) correspond to the intersection point(s) of the line and the parabola.

\begin{cases} y = x + 1 \\ y = x^2 - 2x + 1 \end{cases}

What are all solution(s) $(x,y)$ ?

$(0,1)$ only
$(0,1)$ and $(3,4)$ (correct answer)
$(1,2)$ and $(2,3)$
No solution

y = mx + b

form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system

y = x + 1

and

y = x^2 - 2x + 1

, substitute the linear expression into the quadratic:

x + 1 = x^2 - 2x + 1

, rearrange to

x^2 - 3x = 0

, factor as

x(x - 3) = 0

, so

x = 0

x = 3

; back-substitute to get

y = 1

and

y = 4

, giving points

(0, 1)

and

(3, 4)

. Choice B correctly identifies both intersection points through proper substitution and solving. Remember the transferable strategy: always check the discriminant of the resulting quadratic to predict the number of solutions, and verify each point in both equations to ensure accuracy—you've got this!

Question 12

A ball's height is modeled by $y=-x^2+6x$ and a laser beam is modeled by $y=2x$ . The ball and laser meet at the intersection point(s). Solve the system.

\begin{cases} y = -x^2 + 6x \\ y = 2x \end{cases}

What are all intersection points $(x,y)$ ?

$(0,0)$ and $(4,8)$ (correct answer)
$(0,0)$ and $(2,4)$
$(1,2)$ and $(5,10)$
$(0,2)$ and $(4,8)$

Explanation: This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (parabola) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: solve the linear equation for y (often already in y = mx + b form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system y = -x² + 6x and y = 2x, substitute: -x² + 6x = 2x, rearrange to -x² + 4x = 0, factor -x(x - 4) = 0, x = 0 or 4, y = 0 or 8, giving (0, 0) and (4, 8). Choice A correctly finds both points with proper steps. Fantastic—modeling real-world scenarios like this builds intuition!

Question 13

A ball's height is modeled by $y=-x^2+6x$ and a laser beam is modeled by $y=2x$ . Where do they intersect? (Solve the system; intersection points are the solutions.)

\begin{cases} y = -x^2 + 6x \\ y = 2x \end{cases}

$(0,6)$ and $(4,2)$
$(0,0)$ and $(4,8)$ (correct answer)
$(0,0)$ and $(2,4)$
$(1,2)$ and $(3,6)$

Explanation: This question tests your ability to solve linear-quadratic systems modeling real-world scenarios like a ball's path and a laser beam, finding intersection points. A linear-quadratic system involves a line and a parabola, with solutions as their 0 to 2 intersection points. For

y = -x^2 + 6x

and

y = 2x

, set equal:

-x^2 + 6x = 2x

, so

-x^2 + 4x = 0

x^2 - 4x = 0

x(x - 4) = 0

;

x = 0

4

y = 0

8

, giving

(0, 0)

and

(4, 8)

, both verified. Choice A correctly lists both points from accurate solving. Choice B might arise from factoring errors like

x(x - 2) = 0

, but distribute negatives carefully when rearranging! Apply the steps: substitute, solve quadratic, back-substitute, verify; consider the parabola's vertex for intuition. You're making fantastic progress—keep practicing for modeling success!

Question 14

Solve the linear-quadratic system algebraically (by substitution). The solutions correspond to the intersection point(s) of the line and the parabola.

\begin{cases} y = x + 1 \\ y = x^2 - 2x + 1 \end{cases}

What are all solutions $(x,y)$ to the system?

$(0,1)$ only
$(1,2)$ and $(2,3)$
$(0,1)$ and $(3,4)$ (correct answer)
No real solution

Explanation: This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (parabola) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve like a parabola: the solutions are the intersection points where the line and parabola meet, and there can be 0, 1, or 2 real solutions depending on how they intersect. For the system y = x + 1 and y = x² - 2x + 1, substitute the linear into the quadratic: x + 1 = x² - 2x + 1, rearrange to x² - 3x = 0, factor as x(x - 3) = 0, so x = 0 or x = 3; then y = 1 and y = 4, giving points (0, 1) and (3, 4), which both satisfy the original equations. Choice C correctly identifies both intersection points through accurate substitution and back-substitution. Choice A might tempt if you mistakenly discard x = 3, but remember quadratics often yield two solutions—check both! To master linear-quadratic systems, always substitute the linear expression into the quadratic, solve the resulting quadratic equation, back-substitute to find y-values, and verify each point in both equations. You're doing great—practicing this method will make solving these systems second nature!

Question 15

A ball's height is modeled by $y=-x^2+6x$ and a laser beam is modeled by $y=2x$ . The ball and laser meet at the intersection point(s). Solve the system.

\begin{cases} y = -x^2 + 6x \\ y = 2x \end{cases}

What are all intersection points $(x,y)$ ?

$(0,0)$ and $(2,4)$
$(0,2)$ and $(4,8)$
$(0,0)$ and $(4,8)$ (correct answer)
$(1,2)$ and $(5,10)$

y = -x^2 + 6x

and

y = 2x

, substitute:

-x^2 + 6x = 2x

, rearrange to

-x^2 + 4x = 0

, factor

-x(x - 4) = 0

x = 0

4

y = 0

8

, giving

(0, 0)

and

(4, 8)

. Choice A correctly finds both points with proper steps. Fantastic—modeling real-world scenarios like this builds intuition!

Question 16

A ball's height is modeled by $y=-x^2+6x$ and a laser beam is modeled by $y=2x$ . Where do they intersect? (Solve the system; intersection points are the solutions.)

\begin{cases} y = -x^2 + 6x \\ y = 2x \end{cases}

$(0,0)$ and $(4,8)$ (correct answer)
$(0,0)$ and $(2,4)$
$(1,2)$ and $(3,6)$
$(0,6)$ and $(4,2)$

y = -x^2 + 6x

and

y = 2x

, set equal:

-x^2 + 6x = 2x

, so

-x^2 + 4x = 0

x^2 - 4x = 0

x(x - 4) = 0

;

x = 0

4

y = 0

8

, giving

(0, 0)

and

(4, 8)

, both verified. Choice A correctly lists both points from accurate solving. Choice B might arise from factoring errors like

x(x - 2) = 0

Question 17

Solve the system algebraically. The solution(s) are the intersection point(s) of the line and the parabola.

\begin{cases} y = -x + 1 \\ y = x^2 + 2x - 3 \end{cases}

How many real solutions does the system have?

0 real solutions
1 real solution
2 real solutions (correct answer)
Infinitely many solutions

y = mx + b

form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system

y = -x + 1

and

y = x^2 + 2x - 3

, substitute:

-x + 1 = x^2 + 2x - 3

, rearrange to

x^2 + 3x - 4 = 0

, discriminant

9 + 16 = 25 > 0

, so two real solutions (

x = 1

and

x = -4

). Choice C correctly identifies there are 2 real solutions based on the positive discriminant. Excellent—using the discriminant is a quick way to preview the number without full solving!

Question 18

Find all points where the line intersects the circle (solve the system).

\begin{cases} y = x \\ x^2 + y^2 = 8 \end{cases}

$(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$
$(\sqrt{8},\sqrt{8})$ and $(-\sqrt{8},-\sqrt{8})$
$(2,2)$ and $(-2,-2)$ (correct answer)
$(\sqrt{2},-\sqrt{2})$ and $(-\sqrt{2},\sqrt{2})$

Explanation: This question tests your ability to solve systems of a line and circle to find all intersection points. A linear-quadratic system with a circle can have 0, 1, or 2 intersections based on the line's position. For y = x and

x^2 + y^2 = 8

, substitute:

x^2 + x^2 = 8

2x^2 = 8

x^2 = 4

x = \pm 2

;

y = \pm 2

, giving

(2,2)

and

(-2,-2)

, both valid. Choice A correctly identifies the points from proper substitution. Choice B might result from

\sqrt{8}

instead of

\sqrt{4}

, but solve the quadratic accurately after combining terms! Strategy: substitute linear into circle, simplify, solve for x, find y, verify; discriminant previews the count. Wonderful effort—this will sharpen your geometry-algebra skills!

Question 19

Find the intersection point(s) of the line and the circle by solving the system.

\begin{cases} x^2 + y^2 = 25 \\ y = -3x \end{cases}

What are all solutions $(x,y)$ ?

$(1,-3)$ and $(-1,3)$
$(\pm 2, \mp 6)$
$\left(\frac{5}{\sqrt{10}},-\frac{15}{\sqrt{10}}\right)$ and $\left(-\frac{5}{\sqrt{10}},\frac{15}{\sqrt{10}}\right)$ (correct answer)
$\left(\frac{5}{2},-\frac{15}{2}\right)$ and $\left(-\frac{5}{2},\frac{15}{2}\right)$

Explanation: This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (circle) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: substitute the linear expression into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system x² + y² = 25 and y = -3x, substitute: x² + (-3x)² = 25, so 10x² = 25, x² = 2.5, x = ±√(5/2) = ±5/√10 (rationalized), y = -3x giving (5/√10, -15/√10) and (-5/√10, 15/√10). Choice C correctly finds both intersection points using proper substitution. Great job—circles follow the same method as parabolas!

Question 20

How many real solutions (intersection points) does the linear-quadratic system have?

\begin{cases} y = x^2 + 2x - 3 \\ y = -x + 1 \end{cases} $$​

0 real solutions
Infinitely many solutions
2 real solutions (correct answer)
1 real solution

Explanation: This question tests your ability to determine the number of real solutions in a linear-quadratic system without fully solving, focusing on intersection points of a parabola and line. A linear-quadratic system can have 0, 1, or 2 real intersections based on the discriminant of the resulting quadratic after substitution. For y = x² + 2x - 3 and y = -x + 1, set equal: x² + 2x - 3 = -x + 1, so x² + 3x - 4 = 0; discriminant 9 + 16 = 25 > 0, indicating two real solutions. Choice C correctly identifies two real solutions from the positive discriminant. Choice A might tempt if you miscalculate the discriminant as negative, but double-check arithmetic—b² - 4ac is key! Use the strategy: substitute to form ax² + bx + c = 0, compute discriminant to predict solutions (positive: 2, zero: 1, negative: 0), then solve if needed. Great job exploring this—you'll ace predicting solution counts in no time!