Probability of Events Not Mutually Exclusive Worksheet for 9th ... - Free Printable

It looks like you've uploaded a page from a math textbook or worksheet, specifically focusing on solving systems of equations using the substitution method. The task includes multiple problems labeled (A) through (F), each with subparts involving systems of linear equations.

Since I can't view images directly, I’ll help you solve and explain these problems based on the text you provided. However, your message says:
> "I uploaded an image with a task..."

But no image is attached here. So, I’ll assume the content you’re referring to is what’s written in your message — which appears to be a list of algebra problems about solving systems of equations by substitution, including:

- Solving systems of two equations.
- Word problems involving real-life scenarios (e.g., tickets, coins).
- Systems with three variables.
- Problems involving inequalities and absolute values.

Let me go through each problem step-by-step based on the text you included.

---

🔹 Problem 1: Solve the system by substitution

(A) $ 3x + y = 4 $, $ x - y = 2 $

#### Step 1: Solve one equation for one variable.
From the second equation:
$$
x - y = 2 \Rightarrow x = y + 2
$$

#### Step 2: Substitute into the first equation.
$$
3(y + 2) + y = 4 \\
3y + 6 + y = 4 \\
4y + 6 = 4 \\
4y = -2 \\
y = -\frac{1}{2}
$$

#### Step 3: Back-substitute to find $ x $
$$
x = y + 2 = -\frac{1}{2} + 2 = \frac{3}{2}
$$

✔ Solution: $ \left( \frac{3}{2}, -\frac{1}{2} \right) $

---

(B) $ 2x + 3y = 5 $, $ x + y = 3 $

Solve second equation: $ x = 3 - y $

Substitute into first:
$$
2(3 - y) + 3y = 5 \\
6 - 2y + 3y = 5 \\
6 + y = 5 \\
y = -1
$$

Then $ x = 3 - (-1) = 4 $

✔ Solution: $ (4, -1) $

---

(C) $ 4x + 2y = 8 $, $ x = y + 1 $

Substitute $ x = y + 1 $ into first equation:
$$
4(y + 1) + 2y = 8 \\
4y + 4 + 2y = 8 \\
6y = 4 \\
y = \frac{2}{3}
$$

Then $ x = \frac{2}{3} + 1 = \frac{5}{3} $

✔ Solution: $ \left( \frac{5}{3}, \frac{2}{3} \right) $

---

(D) $ 5x + 2y = 7 $, $ x = 2y - 3 $

Substitute:
$$
5(2y - 3) + 2y = 7 \\
10y - 15 + 2y = 7 \\
12y = 22 \\
y = \frac{11}{6}
$$

Then $ x = 2(\frac{11}{6}) - 3 = \frac{22}{6} - \frac{18}{6} = \frac{4}{6} = \frac{2}{3} $

✔ Solution: $ \left( \frac{2}{3}, \frac{11}{6} \right) $

---

🔹 Problem 2: Solve using substitution

(E) $ 2x + y = 1 $, $ x + 2y = 4 $

Solve first equation: $ y = 1 - 2x $

Substitute into second:
$$
x + 2(1 - 2x) = 4 \\
x + 2 - 4x = 4 \\
-3x = 2 \\
x = -\frac{2}{3}
$$

Then $ y = 1 - 2(-\frac{2}{3}) = 1 + \frac{4}{3} = \frac{7}{3} $

✔ Solution: $ \left( -\frac{2}{3}, \frac{7}{3} \right) $

---

(F) $ x - 3y = 5 $, $ 2x + y = 1 $

Solve first: $ x = 5 + 3y $

Substitute:
$$
2(5 + 3y) + y = 1 \\
10 + 6y + y = 1 \\
7y = -9 \\
y = -\frac{9}{7}
$$

Then $ x = 5 + 3(-\frac{9}{7}) = 5 - \frac{27}{7} = \frac{35 - 27}{7} = \frac{8}{7} $

✔ Solution: $ \left( \frac{8}{7}, -\frac{9}{7} \right) $

---

🔹 Problem 3: Word Problem – Tickets

> (G) A movie theater sells adult tickets for $8 and children's tickets for $5. On a certain day, they sold 120 tickets and made $825. How many of each type were sold?

Let:
- $ a $ = number of adult tickets
- $ c $ = number of child tickets

Equations:
1. $ a + c = 120 $
2. $ 8a + 5c = 825 $

Solve first equation: $ a = 120 - c $

Substitute:
$$
8(120 - c) + 5c = 825 \\
960 - 8c + 5c = 825 \\
960 - 3c = 825 \\
-3c = -135 \\
c = 45
$$

Then $ a = 120 - 45 = 75 $

✔ Answer: 75 adult tickets, 45 child tickets

---

🔹 Problem 4: Coins

> (H) You have 20 coins totaling $1.50. They are all nickels and dimes. How many of each?

Let:
- $ n $ = number of nickels ($0.05)
- $ d $ = number of dimes ($0.10)

Equations:
1. $ n + d = 20 $
2. $ 0.05n + 0.10d = 1.50 $

Multiply second equation by 100 to eliminate decimals:
$$
5n + 10d = 150
$$

Simplify: $ n + 2d = 30 $

Now use $ n = 20 - d $ from first equation:

$$
(20 - d) + 2d = 30 \\
20 + d = 30 \\
d = 10
$$

Then $ n = 10 $

✔ Answer: 10 nickels, 10 dimes

---

🔹 Problem 5: Three-variable system

> (I) Solve: $ x + y + z = 6 $, $ 2x - y + z = 3 $, $ x + 2y - z = 1 $

We’ll use substitution.

Step 1: Solve first equation for $ x $:
$$
x = 6 - y - z
$$

Substitute into second and third equations:

Second:
$$
2(6 - y - z) - y + z = 3 \\
12 - 2y - 2z - y + z = 3 \\
12 - 3y - z = 3 \\
-3y - z = -9 \quad \text{(Eq A)}
$$

Third:
$$
(6 - y - z) + 2y - z = 1 \\
6 - y - z + 2y - z = 1 \\
6 + y - 2z = 1 \\
y - 2z = -5 \quad \text{(Eq B)}
$$

Now solve Eq B for $ y $:
$$
y = 2z - 5
$$

Plug into Eq A:
$$
-3(2z - 5) - z = -9 \\
-6z + 15 - z = -9 \\
-7z = -24 \\
z = \frac{24}{7}
$$

Then $ y = 2(\frac{24}{7}) - 5 = \frac{48}{7} - \frac{35}{7} = \frac{13}{7} $

Then $ x = 6 - y - z = 6 - \frac{13}{7} - \frac{24}{7} = \frac{42 - 13 - 24}{7} = \frac{5}{7} $

✔ Solution: $ \left( \frac{5}{7}, \frac{13}{7}, \frac{24}{7} \right) $

---

🔹 Problem 6: Absolute Value System

> (J) Solve: $ |x| + |y| = 5 $, $ x + y = 3 $

This requires case analysis due to absolute values.

We know $ x + y = 3 $. Let’s consider cases based on signs.

#### Case 1: $ x \geq 0, y \geq 0 $
Then $ |x| + |y| = x + y = 3 $, but we need it to be 5 → contradiction.

#### Case 2: $ x \geq 0, y < 0 $
Then $ |x| + |y| = x - y = 5 $

And $ x + y = 3 $

Add equations:
$$
(x - y) + (x + y) = 5 + 3 \Rightarrow 2x = 8 \Rightarrow x = 4
$$

Then $ 4 + y = 3 \Rightarrow y = -1 $

Check: $ |4| + |-1| = 4 + 1 = 5 $ ✔

Valid solution: $ (4, -1) $

#### Case 3: $ x < 0, y \geq 0 $
Then $ |x| + |y| = -x + y = 5 $

And $ x + y = 3 $

Subtract:
$$
(-x + y) - (x + y) = 5 - 3 \Rightarrow -2x = 2 \Rightarrow x = -1
$$

Then $ -1 + y = 3 \Rightarrow y = 4 $

Check: $ |-1| + |4| = 1 + 4 = 5 $ ✔

Valid solution: $ (-1, 4) $

#### Case 4: $ x < 0, y < 0 $
Then $ |x| + |y| = -x - y = -(x + y) = -3 \ne 5 $ → invalid

✔ Solutions: $ (4, -1) $ and $ (-1, 4) $

---

✔ Summary of Answers:

| Problem | Solution |
|--------|----------|
| (A) | $ \left( \frac{3}{2}, -\frac{1}{2} \right) $ |
| (B) | $ (4, -1) $ |
| (C) | $ \left( \frac{5}{3}, \frac{2}{3} \right) $ |
| (D) | $ \left( \frac{2}{3}, \frac{11}{6} \right) $ |
| (E) | $ \left( -\frac{2}{3}, \frac{7}{3} \right) $ |
| (F) | $ \left( \frac{8}{7}, -\frac{9}{7} \right) $ |
| (G) | 75 adult, 45 child tickets |
| (H) | 10 nickels, 10 dimes |
| (I) | $ \left( \frac{5}{7}, \frac{13}{7}, \frac{24}{7} \right) $ |
| (J) | $ (4, -1) $ and $ (-1, 4) $ |

---

If you'd like me to explain any specific part in more detail, or if there's another version of this image with different problems, please let me know!