M3 - Systems of Equations (quadratic & linear) - Free Printable

Let's solve each of the linear-quadratic systems algebraically. The goal is to find the points of intersection between a quadratic equation (parabola) and a linear equation (line). We do this by substituting the linear expression into the quadratic equation, then solving for $ x $. After finding $ x $, we substitute back to get $ y $.

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1.

$$
y = x^2 + 3x - 5 \\
y = x + 3
$$

Set the right-hand sides equal:
$$
x^2 + 3x - 5 = x + 3
$$
Subtract $ x + 3 $ from both sides:
$$
x^2 + 2x - 8 = 0
$$
Factor:
$$
(x + 4)(x - 2) = 0
\Rightarrow x = -4 \text{ or } x = 2
$$

Now find $ y $ using $ y = x + 3 $:

- If $ x = -4 $, $ y = -4 + 3 = -1 $
- If $ x = 2 $, $ y = 2 + 3 = 5 $

✔ Solutions: $ (-4, -1), (2, 5) $

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2.

$$
y = x^2 - 4x + 6 \\
y = x + 2
$$

Set equal:
$$
x^2 - 4x + 6 = x + 2
\Rightarrow x^2 - 5x + 4 = 0
$$
Factor:
$$
(x - 1)(x - 4) = 0
\Rightarrow x = 1 \text{ or } x = 4
$$

Use $ y = x + 2 $:

- $ x = 1 \Rightarrow y = 3 $
- $ x = 4 \Rightarrow y = 6 $

✔ Solutions: $ (1, 3), (4, 6) $

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3.

$$
y = x^2 - 10x + 14 \\
y = 7x - 16
$$

Set equal:
$$
x^2 - 10x + 14 = 7x - 16
\Rightarrow x^2 - 17x + 30 = 0
$$
Factor:
$$
(x - 15)(x - 2) = 0
\Rightarrow x = 15 \text{ or } x = 2
$$

Use $ y = 7x - 16 $:

- $ x = 15 \Rightarrow y = 105 - 16 = 89 $
- $ x = 2 \Rightarrow y = 14 - 16 = -2 $

✔ Solutions: $ (15, 89), (2, -2) $

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4.

$$
y = x^2 - 24 \\
y = x - 12
$$

Set equal:
$$
x^2 - 24 = x - 12
\Rightarrow x^2 - x - 12 = 0
$$
Factor:
$$
(x - 4)(x + 3) = 0
\Rightarrow x = 4 \text{ or } x = -3
$$

Use $ y = x - 12 $:

- $ x = 4 \Rightarrow y = -8 $
- $ x = -3 \Rightarrow y = -15 $

✔ Solutions: $ (4, -8), (-3, -15) $

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5.

$$
y = x^2 - 8x - 12 \\
y = 4x + 8
$$

Set equal:
$$
x^2 - 8x - 12 = 4x + 8
\Rightarrow x^2 - 12x - 20 = 0
$$
Factor:
Try: $ (x - 10)(x + 2) = x^2 - 8x - 20 $? No.

Discriminant: $ d = 144 + 80 = 224 $ → not perfect square.

Use quadratic formula:
$$
x = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(-20)}}{2}
= \frac{12 \pm \sqrt{144 + 80}}{2}
= \frac{12 \pm \sqrt{224}}{2}
= \frac{12 \pm 4\sqrt{14}}{2}
= 6 \pm 2\sqrt{14}
$$

So,
- $ x = 6 + 2\sqrt{14} $, $ y = 4x + 8 = 4(6 + 2\sqrt{14}) + 8 = 24 + 8\sqrt{14} + 8 = 32 + 8\sqrt{14} $
- $ x = 6 - 2\sqrt{14} $, $ y = 4(6 - 2\sqrt{14}) + 8 = 24 - 8\sqrt{14} + 8 = 32 - 8\sqrt{14} $

✔ Solutions: $ (6 + 2\sqrt{14}, 32 + 8\sqrt{14}), (6 - 2\sqrt{14}, 32 - 8\sqrt{14}) $

---

6.

$$
y = x^2 + 6x + 3 \\
y = 3x - 7
$$

Set equal:
$$
x^2 + 6x + 3 = 3x - 7
\Rightarrow x^2 + 3x + 10 = 0
$$

Discriminant: $ 9 - 40 = -31 < 0 $

✘ No real solutions

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7.

$$
y = x^2 - 9x - 18 \\
y = x + 3
$$

Set equal:
$$
x^2 - 9x - 18 = x + 3
\Rightarrow x^2 - 10x - 21 = 0
$$
Factor:
$$
(x - 11)(x + 1) = 0 \quad \text{(Check: } 11 \cdot (-1) = -11, no)}
$$

Wait: $ x^2 - 10x - 21 $ — factors: need two numbers that multiply to -21 and add to -10.

Try: $ -11 $ and $ +1 $: $ -11 + 1 = -10 $, $ -11 \cdot 1 = -11 $ → no

Try: $ -7 $ and $ +3 $: $ -7+3=-4 $, no

$ -14 $ and $ +3 $: too big

Use quadratic formula:
$$
x = \frac{10 \pm \sqrt{100 + 84}}{2} = \frac{10 \pm \sqrt{184}}{2} = \frac{10 \pm 2\sqrt{46}}{2} = 5 \pm \sqrt{46}
$$

Then $ y = x + 3 $:

- $ x = 5 + \sqrt{46} \Rightarrow y = 8 + \sqrt{46} $
- $ x = 5 - \sqrt{46} \Rightarrow y = 8 - \sqrt{46} $

✔ Solutions: $ (5 + \sqrt{46}, 8 + \sqrt{46}), (5 - \sqrt{46}, 8 - \sqrt{46}) $

---

8.

$$
y = x^2 + 6x + 10 \\
y = -2x - 6
$$

Set equal:
$$
x^2 + 6x + 10 = -2x - 6
\Rightarrow x^2 + 8x + 16 = 0
$$
Factor:
$$
(x + 4)^2 = 0 \Rightarrow x = -4
$$

Then $ y = -2(-4) - 6 = 8 - 6 = 2 $

✔ Solution: $ (-4, 2) $ (one solution — tangent point)

---

9.

$$
y = x^2 + 8x + 16 \\
y = x + 6
$$

Set equal:
$$
x^2 + 8x + 16 = x + 6
\Rightarrow x^2 + 7x + 10 = 0
$$
Factor:
$$
(x + 5)(x + 2) = 0 \Rightarrow x = -5 \text{ or } x = -2
$$

Use $ y = x + 6 $:

- $ x = -5 \Rightarrow y = 1 $
- $ x = -2 \Rightarrow y = 4 $

✔ Solutions: $ (-5, 1), (-2, 4) $

---

10.

$$
y = x^2 - 3x - 6 \\
y = x + 6
$$

Set equal:
$$
x^2 - 3x - 6 = x + 6
\Rightarrow x^2 - 4x - 12 = 0
$$
Factor:
$$
(x - 6)(x + 2) = 0 \Rightarrow x = 6 \text{ or } x = -2
$$

Use $ y = x + 6 $:

- $ x = 6 \Rightarrow y = 12 $
- $ x = -2 \Rightarrow y = 4 $

✔ Solutions: $ (6, 12), (-2, 4) $

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✔ Final Answers Summary:

| Problem | Solutions |
|--------|-----------|
| 1 | $ (-4, -1), (2, 5) $ |
| 2 | $ (1, 3), (4, 6) $ |
| 3 | $ (15, 89), (2, -2) $ |
| 4 | $ (4, -8), (-3, -15) $ |
| 5 | $ (6 + 2\sqrt{14}, 32 + 8\sqrt{14}), (6 - 2\sqrt{14}, 32 - 8\sqrt{14}) $ |
| 6 | No real solutions |
| 7 | $ (5 + \sqrt{46}, 8 + \sqrt{46}), (5 - \sqrt{46}, 8 - \sqrt{46}) $ |
| 8 | $ (-4, 2) $ |
| 9 | $ (-5, 1), (-2, 4) $ |
| 10 | $ (6, 12), (-2, 4) $ |

Let me know if you'd like these graphed or explained further!