Problem 6 · AMC 8 Stretch
Core
Algebra & Patterns
logical-reasoningrecognizing-false-rules
Many students freeze on \(5x=3x\) because there seems to be 'nothing' on the right. Solve \(5x=3x\) using the same first step you would use for \(5x=3x+6\).
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Answer: x = 0
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Hint 1 of 4
Do the very same first step you would do for \(5x=3x+6\): get all the \(x\)'s onto one side. Don't be thrown off by the missing number.
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Hint 2 of 4
Subtract \(3x\) from both sides. What is left on the left? What is left on the right?
Still stuck? Show hint 3 →
Hint 3 of 4
You get \(2x=0\). The right side is the NUMBER zero — a real number you can work with, not 'nothing.'
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Approach: Treat zero as a number and solve normally
- Use the same move as for \(5x=3x+6\): subtract \(3x\) from both sides.
- \(5x-3x=3x-3x\) gives \(2x=0\). The right side is the NUMBER zero, not 'nothing.'
- Divide both sides by \(2\): \(x=0\).
- So \(x=0\). The equation was never harder than the first one — it only felt that way because \(0\) can look like 'nothing.'
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