Solve for the variable in these pre-algebra equations using inverse operations.

I'll solve each equation step by step, showing my work for all 10 problems.

Problem 1: x + y + 9 = 18
Wait, this appears to be "x × y + 9 = 18" or similar notation. Let me interpret the notation carefully. Looking at the pattern, it seems like these use a special notation where numbers are combined with variables. Let me work through each one:

Actually, looking more carefully at the notation used (like "1xy", "5×(8-y)", etc.), I need to interpret what these mean in pre-algebra context. The most likely interpretation is that these are multiplication expressions where adjacent symbols mean multiplication.

Let me solve each problem:

Problem 1: If this means x·y + 9 = 18, then x·y = 9. But we have two unknowns. This doesn't make sense for finding a single variable.

Looking again at the format - perhaps the first number is a coefficient? So "1xy" might mean "1 × x × y"? That still gives us xy = 9.

Wait - let me reconsider. In some pre-algebra contexts, when they write something like "1xy" they might mean a specific value for x times y. But typically you'd solve for ONE variable.

Let me try another interpretation: Perhaps these equations have a typo or unusual notation. Let me assume standard algebraic notation where:
- Numbers next to parentheses mean multiplication
- We're solving for the variable shown

Let me re-examine: Looking at problem 2: "5×(8-y) = 0" - this is clear notation. So perhaps problem 1 is meant to be read differently.

Actually, I think the notation "1xy" might mean "x" with subscript or it's just poorly formatted and should be "x + 9 = 18". Let me try that interpretation for odd-numbered problems and see if it makes sense.

Let me try: Problem 1 as "x + 9 = 18" → x = 9 ✓
Problem 3 as "12x + 7 = 151" → 12x = 144 → x = 12 ✓
Problem 5 as "(y5)² = 3600" - hmm, this is unclear

Let me try yet another approach. What if the first digit is actually part of a multi-digit coefficient? So:
- Problem 1: Could be incomplete or "x + y + 9 = 18"
- Problem 3: "12x + y + 7 = 151" or "12xy + 7 = 151"?

Given the confusion, let me look at clearer problems first:

Problem 2: 5 × (8 - y) = 0
5(8 - y) = 0
Divide both sides by 5: 8 - y = 0
y = 8 ✓

Problem 4: 3xy + 5 = 14 (assuming this means 3·x·y + 5 = 14)
This has two variables. Unless... maybe it's "3x + y + 5 = 14"? Or perhaps "3x·y" where we solve for one?

Let me try: 3x + 5 = 14 (ignoring y)
3x = 9
x = 3

Or if it's 3y + 5 = 14:
3y = 9
y = 3

Problem 6: 2 × (10 + y) = 44
2(10 + y) = 44
Divide by 2: 10 + y = 22
y = 12 ✓

Problem 7: 11 × (12 - y) = 11
11(12 - y) = 11
Divide by 11: 12 - y = 1
y = 11 ✓

Problem 8: 7xy + 8 = 71 (or 7x + y + 8 = 71?)
If 7y + 8 = 71:
7y = 63
y = 9 ✓

Problem 9: 5xy + 8 = 53 (or 5x + y + 8 = 53?)
If 5y + 8 = 53:
5y = 45
y = 9 ✓

Problem 10: (y1)² = 64 - This notation is unclear. Could be (y·1)² = 64, meaning y² = 64, so y = 8 or y = -8

For problem 5: (y5)² = 3600 - Could be (y·5)² = 3600, meaning 25y² = 3600, so y² = 144, y = 12 or y = -12

Let me verify these interpretations make sense and rec