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

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

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

Every worksheet problem on systems of linear and quadratic equations is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where forgetting to find the y-values can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

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

Nonlinear systems answers: exercises 1 to 10

Exercise 1 – Check a point

First equation: $2^{2} + 1 = 5$, true. Second: $3(2) – 1 = 5$, true.

Yes, (2, 5) satisfies both equations

Exercise 2 – Substitution

Substitute $y = x + 6$ into $y = x^{2}$: $x^{2} = x + 6$.

Move everything to one side: $x^{2} – x – 6 = 0$.

Solve: $x = -2\text{ or }x = 3$.

Find each y with the linear equation: $y = x + 6 = 4$ when $x = -2$; $y = x + 6 = 9$ when $x = 3$.

Solutions: $(-2, 4), (3, 9)$

Exercise 3 – Parabola and line

Substitute $y = x – 1$ into $y = x^{2} – 4x + 3$: $x^{2} – 4x + 3 = x – 1$.

Move everything to one side: $x^{2} – 5x + 4 = 0$.

Solve: $x = 1\text{ or }x = 4$.

Find each y with the linear equation: $y = x – 1 = 0$ when $x = 1$; $y = x – 1 = 3$ when $x = 4$.

Solutions: $(1, 0), (4, 3)$

Exercise 4 – Read the graph

Intersections (-4, 5) and (1, 0).

Substitute $y = -x + 1$ into $y = x^{2} + 2x – 3$: $x^{2} + 2x – 3 = -x + 1$.

Move everything to one side: $x^{2} + 3x – 4 = 0$.

Solve: $x = -4\text{ or }x = 1$.

Find each y with the linear equation: $y = -x + 1 = 5$ when $x = -4$; $y = -x + 1 = 0$ when $x = 1$.

Solutions: $(-4, 5), (1, 0)$

Exercise 5 – Tangent line

Substitute $y = 2x$ into $y = x^{2} + 1$: $x^{2} + 1 = 2x$.

Move everything to one side: $x^{2} – 2x + 1 = 0$.

The equation has one repeated solution: $x = 1$.

Find each y with the linear equation: $y = 2x = 2$ when $x = 1$.

Solutions: $(1, 2)$

Because there is exactly one point of intersection, the line is tangent to the parabola.

Exercise 6 – No real solution

$x^{2} + 3 = 2x$ gives $x^{2} – 2x + 3 = 0$.

$D = (-2)^{2} – 4(1)(3) = -8 < 0$.

No real solutions: the line does not meet the parabola

Exercise 7 – Line in standard form

Rewrite the line as $y = -x + 5$.

Substitute $y = -x + 5$ into $y = x^{2} – 1$: $x^{2} – 1 = -x + 5$.

Move everything to one side: $x^{2} + x – 6 = 0$.

Solve: $x = -3\text{ or }x = 2$.

Find each y with the linear equation: $y = -x + 5 = 8$ when $x = -3$; $y = -x + 5 = 3$ when $x = 2$.

Solutions: $(-3, 8), (2, 3)$

Exercise 8 – Irrational solutions

$2x^{2} – 3x – 5 = x + 3$ gives $2x^{2} – 4x – 8 = 0$, so $x^{2} – 2x – 4 = 0$.

$x = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}$.

Then $y = x + 3 = 4 \pm \sqrt{5}$.

$(1 + \sqrt{5}, 4 + \sqrt{5})$ and $(1 – \sqrt{5}, 4 – \sqrt{5})$

Exercise 9 – Solve and graph

Intersections (-3, 5) and (2, 0).

Substitute $y = -x + 2$ into $y = x^{2} – 4$: $x^{2} – 4 = -x + 2$.

Move everything to one side: $x^{2} + x – 6 = 0$.

Solve: $x = -3\text{ or }x = 2$.

Find each y with the linear equation: $y = -x + 2 = 5$ when $x = -3$; $y = -x + 2 = 0$ when $x = 2$.

Solutions: $(-3, 5), (2, 0)$

Exercise 10 – Circle and line

$x^{2} + (x + 1)^{2} = 25$ gives $2x^{2} + 2x – 24 = 0$, so $x^{2} + x – 12 = 0$.

$(x + 4)(x – 3) = 0$. $y = x + 1$: $y = -3$ or $y = 4$.

$(3, 4)$ and $(-4, -3)$

Exercises 11 and up

Exercise 11 – Circle and line through the origin

$x^{2} + 9x^{2} = 10$, so $x^{2} = 1$ and $x = \pm 1$.

$y = 3x = \pm 3$.

$(1, 3)$ and $(-1, -3)$

Exercise 12 – Tangent to a circle

Tangent at (-2, 2).

$x^{2} + (x + 4)^{2} = 8$ gives $2x^{2} + 8x + 8 = 0$, so $x^{2} + 4x + 4 = 0$.

$(x + 2)^{2} = 0$, so $x = -2$ only and $y = 2$.

One intersection point (-2, 2), so the line is tangent

Exercise 13 – Count the intersections

Completed table.

$x^{2} – 4x + 4 = 0$: $D = 0$. $x^{2} – 2x + 5 = 0$: $D = 4 – 20 = -16$. $x^{2} – x – 2 = 0$: $D = 1 + 8 = 9$.

1, 0, and 2 intersection points

Exercise 14 – Find the parameter

$x^{2} = 2x + k$ gives $x^{2} – 2x – k = 0$ with $D = 4 + 4k$.

One point when $D = 0$: $k = -1$. Two points when $D > 0$: $k > -1$. None when $k < -1$.

One: k = -1; two: k > -1; none: k < -1

Exercise 15 – Perimeter and area

$y = 15 – x$, so $x(15 – x) = 56$ and $x^{2} – 15x + 56 = 0$.

$(x – 7)(x – 8) = 0$, so $x = 7$ or $8$, and $y = 8$ or $7$.

7 meters by 8 meters

Exercise 16 – Two numbers

$x + y = 12$, $xy = 32$: $x(12 – x) = 32$ gives $x^{2} – 12x + 32 = 0$.

$(x – 4)(x – 8) = 0$.

4 and 8

Exercise 17 – Ball and ramp

Meeting points (1, 5) and (4, 8).

$-x^{2} + 6x = x + 4$ gives $x^{2} – 5x + 4 = 0$, so $(x – 1)(x – 4) = 0$.

$y = x + 4$: $5$ and $8$.

At (1, 5) and (4, 8)

Exercise 18 – Break-even

Break-even at x = 3 and x = 9.

$-x^{2} + 14x = 2x + 27$ gives $x^{2} – 12x + 27 = 0$, so $(x – 3)(x – 9) = 0$.

At $x = 3$: $R = C = 33$. At $x = 9$: $R = C = 45$.

300 items and 900 items ($3,300 and $4,500)

Exercise 19 – Find the error

A solution of a system is an ordered pair. He must find each $y$: $y = 3 + 6 = 9$ and $y = -2 + 6 = 4$.

$(3, 9)$ and $(-2, 4)$

Exercise 20 – Write a system

A line and a parabola can meet in 2, 1, or 0 points.

$(1, 0)$: $1 – 4 + 3 = 0$ and $1 – 1 = 0$. $(4, 3)$: $16 – 16 + 3 = 3$ and $4 – 1 = 3$.

Both points satisfy both equations, so the line crosses the parabola twice.

Two solutions, like the left panel

Key facts behind this nonlinear systems 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 answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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