Worksheet: Order of Operations and Grouping Symbols | Pre-Algebra ... - Free Printable

It looks like you've uploaded a worksheet titled "Factorials Worksheet" with several problems involving factorial expressions and solving equations. Since I can't view images directly, I’ll assume the content of the worksheet is as described in your text.

Let me analyze and solve each problem based on the information provided.

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Problem 1:

Solve for \( x \):

#### A) \( 3! = x \)

We know:
\[
3! = 3 \times 2 \times 1 = 6
\]
So,
\[
x = 6
\]

✔ Answer: \( x = 6 \)

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#### B) \( 3! + 5 = x - 2 \)

First compute \( 3! = 6 \), so:
\[
6 + 5 = x - 2 \\
11 = x - 2 \\
x = 13
\]

✔ Answer: \( x = 13 \)

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#### C) \( 5! = x \cdot 4! \)

We know:
\[
5! = 120, \quad 4! = 24
\]
So:
\[
120 = x \cdot 24 \\
x = \frac{120}{24} = 5
\]

✔ Answer: \( x = 5 \)

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#### D) \( (2!)^2 - 6 = x + 1 \)

Compute:
\[
2! = 2 \Rightarrow (2!)^2 = 2^2 = 4
\]
Then:
\[
4 - 6 = x + 1 \\
-2 = x + 1 \\
x = -3
\]

✔ Answer: \( x = -3 \)

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#### E) \( 7! = x \cdot (2! - 2) \)

Note: \( 2! = 2 \), so \( 2! - 2 = 2 - 2 = 0 \)

So:
\[
7! = x \cdot 0 = 0
\]
But \( 7! = 5040 \neq 0 \), so this equation has no solution.

⚠️ Answer: No solution (undefined, division by zero if solving for \( x \))

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#### F) \( 9! = 4! + 5! + x \)

Compute:
\[
9! = 362880 \\
4! = 24, \quad 5! = 120
\]
So:
\[
362880 = 24 + 120 + x \\
362880 = 144 + x \\
x = 362880 - 144 = 362736
\]

✔ Answer: \( x = 362736 \)

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#### G) \( 1 + 2! + 3! + 4! = x \)

Compute:
\[
1 + 2 + 6 + 24 = 33
\]
So:
\[
x = 33
\]

✔ Answer: \( x = 33 \)

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#### H) \( 2! + 3! + 4! = x \)

\[
2 + 6 + 24 = 32
\]
So:
\[
x = 32
\]

✔ Answer: \( x = 32 \)

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#### I) \( 2(6!) + 3! = x + 9 \)

Compute:
\[
6! = 720 \Rightarrow 2 \cdot 720 = 1440 \\
3! = 6
\]
So:
\[
1440 + 6 = x + 9 \\
1446 = x + 9 \\
x = 1437
\]

✔ Answer: \( x = 1437 \)

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#### J) \( 3! + 3 - (5!) + 5!(3) = 6! \)

Wait — let’s check this carefully. The expression is:

\[
3! + 3 - (5!) + 5!(3) = 6!
\]

But note: This is an equation, not a "solve for x". Let's evaluate both sides.

Left-hand side:
\[
3! = 6 \\
5! = 120 \\
5!(3) = 120 \cdot 3 = 360
\]
So:
\[
6 + 3 - 120 + 360 = (9) + (240) = 249
\]

Right-hand side:
\[
6! = 720
\]

So:
\[
249 = 720 \quad \text{✘ False}
\]

This equation is not true.

⚠️ Answer: The statement is false.

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#### K) \( 2!(3!) + 2! + 3!(5) = 6! \)

Evaluate left-hand side:
\[
2! = 2, \quad 3! = 6
\]
So:
\[
2 \cdot 6 = 12 \\
2! = 2 \\
3!(5) = 6 \cdot 5 = 30
\]
Add:
\[
12 + 2 + 30 = 44
\]
Right-hand side: \( 6! = 720 \)

So:
\[
44 = 720 \quad \text{✘ False}
\]

⚠️ Answer: The statement is false.

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✔ Summary of Answers:

| Problem | Answer |
|--------|--------|
| A) \( 3! = x \) | \( x = 6 \) |
| B) \( 3! + 5 = x - 2 \) | \( x = 13 \) |
| C) \( 5! = x \cdot 4! \) | \( x = 5 \) |
| D) \( (2!)^2 - 6 = x + 1 \) | \( x = -3 \) |
| E) \( 7! = x \cdot (2! - 2) \) | No solution |
| F) \( 9! = 4! + 5! + x \) | \( x = 362736 \) |
| G) \( 1 + 2! + 3! + 4! = x \) | \( x = 33 \) |
| H) \( 2! + 3! + 4! = x \) | \( x = 32 \) |
| I) \( 2(6!) + 3! = x + 9 \) | \( x = 1437 \) |
| J) \( 3! + 3 - 5! + 5!(3) = 6! \) | False |
| K) \( 2!(3!) + 2! + 3!(5) = 6! \) | False |

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