Algebra 2 Systems of Linear and Quadratic Equations Test Solutions and Scoring (PDF)

Algebra 2 math test solutions: Systems of Linear and Quadratic Equations – Maths PDF

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This nonlinear systems test answers for Algebra 2 covers systems of linear and quadratic equations (Common Core A-REI.C.7), with a free PDF to print.

Use this answer key and scoring guide to grade the test on systems of linear and quadratic equations (A-REI.C.7). Each solution shows the full reasoning, and the scoring note explains how many points to award for correct setup, correct work, and a correct final answer. Partial credit rewards a sound method even when a small arithmetic slip changes the result.

Algebra 2 · Common Core State Standards: A-REI.C.7

Nonlinear systems test answers: worked solutions

Question 1 – Check a solution [2 pts]

First: \(9 – 12 + 5 = 2\), true. Second: \(3 – 1 = 2\), true.

Yes

Scoring: 1 point for each equation checked.

Question 2 – Substitution [3 pts]

\(x^{2} + x – 4 = x + 1\) gives \(x^{2} = 5\).

\(x = \pm\sqrt{5}\) and \(y = x + 1 = 1 \pm \sqrt{5}\).

\((\sqrt{5}, 1 + \sqrt{5})\) and \((-\sqrt{5}, 1 – \sqrt{5})\)

Scoring: 1 point for the substituted equation, 1 point for the x-values, 1 point for the y-values.

Question 3 – Graph and algebra [3 pts]

Intersection points.

Substitute \(y = x – 3\) into \(y = x^{2} – 2x – 3\): \(x^{2} – 2x – 3 = x – 3\).

Move everything to one side: \(x^{2} – 3x = 0\).

Solve: \(x = 0\text{ or }x = 3\).

Find each y with the linear equation: \(y = x – 3 = -3\) when \(x = 0\); \(y = x – 3 = 0\) when \(x = 3\).

Solutions: \((0, -3), (3, 0)\)

Scoring: 1 point for the equation, 1 point for the x-values, 1 point for the points.

Question 4 – Circle and line [2 pts]

\(x^{2} + (x – 2)^{2} = 10\) gives \(2x^{2} – 4x – 6 = 0\), so \(x^{2} – 2x – 3 = 0\).

\((x – 3)(x + 1) = 0\). \(y = x – 2\): \(1\) or \(-3\).

\((3, 1)\) and \((-1, -3)\)

Scoring: 1 point for the equation, 1 point for both points.

Question 5 – Number of intersections [2 pts]

\(x^{2} – 4 = k\) gives \(x^{2} = k + 4\). Two real solutions need \(k + 4 > 0\).

\(k > -4\)

Scoring: 1 point for the equation, 1 point for the condition.

Question 6 – Break-even [2 pts]

Break-even points.

\(-x^{2} + 16x = 4x + 20\) gives \(x^{2} – 12x + 20 = 0\), so \((x – 2)(x – 10) = 0\).

x = 2 and x = 10

Scoring: 1 point for the equation, 1 point for the solutions.

Question 7 – Two numbers [2 pts]

\(x – y = 4\), so \(x = y + 4\). \(y(y + 4) = 45\) gives \(y^{2} + 4y – 45 = 0\), \((y + 9)(y – 5) = 0\).

\(y = 5\), \(x = 9\) or \(y = -9\), \(x = -5\).

9 and 5, or -5 and -9

Scoring: 1 point for the equation, 1 point for both pairs.

Question 8 – Tangent line [2 pts]

\(x^{2} = 4x – 4\) gives \(x^{2} – 4x + 4 = 0\), so \((x – 2)^{2} = 0\). There is exactly one solution, \(x = 2\), with \(y = 4\).

Tangent at (2, 4)

Scoring: 1 point for the equation, 1 point for the conclusion with the point.

Question 9 – Find the error [2 pts]

\(x^{2} – 1 = 3\) gives \(x^{2} = 4\), so \(x = \pm 2\). Maya lost the negative solution.

\((2, 3)\) and \((-2, 3)\)

Scoring: 1 point for finding both x-values, 1 point for the points.

Total score

Total

20 points

Key facts behind this nonlinear systems test answers

Key facts

Substitute the line into the quadratic to get one equation in x.
Solve the resulting quadratic; then find y for each x.
D > 0: two intersection points; D = 0: one (tangent); D < 0: none.
A solution is an ordered pair (x, y).
The method also works for a line and a circle.

How to use this nonlinear systems test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

Download the nonlinear systems test answers as a PDF with the button below and keep it with your notes.

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