Depth · one concept from first principles · step 2 of 4
Let the sign decide the exponent's parity
In the equation , the numbers , , and are natural numbers and . Find .
ExpressionsNumber system
Your answer
How to solve
Strategy Convert to Algebra — The sign of the left side depends only on whether a is even or odd, so settle a first by matching signs. With a known, simplify the left side and match the exponents of x and y.
1STEP 1
Look at the sign first
Only (-x³/y)^a can bring a minus sign, since the other two have none; the left coefficient is (-1)^a÷1/4.
(-1)^a×4=4 → (-1)^a=1
A negative number to an odd power stays negative; to an even power it turns positive.
7.NS.A.2Eliminate PossibilitiesThe left side's coefficient is positive only when , that is, when is even.
Why?
The middle factor has no minus sign, and the last one is a square, where the two minus signs multiply to plus, so neither can make the result negative.
🧱Signs of products and quotientsMultiplying by a negative number flips a number to the other side of zero, so two flips land back on the positive side: for two nonzero numbers, a product or quotient with like signs is positive and one with unlike signs is negative (zero itself has no sign).
Why?
is copies of multiplied together, and each pair of s multiplies to .
Why?
A power counts how many times a factor is used.
🧱An exponent counts repeated multiplicationAn exponent records only how many times a factor is used, so multiplying two powers of the same base just adds those counts together.
Why?
Two negatives multiply to a positive, so the s cancel in pairs.
🧱Signs of products and quotientsMultiplying by a negative number flips a number to the other side of zero, so two flips land back on the positive side: for two nonzero numbers, a product or quotient with like signs is positive and one with unlike signs is negative (zero itself has no sign).
Why?
The pairs use up all the s exactly when is even; an odd leaves one over.
🧱Even and odd behave predictablyTwice any whole number is even and adding an odd amount always swaps even with odd, so a total's parity is settled by its parts before anything is computed.
2STEP 2
Pin down a
a must be even and 1 ≤ a ≤ 3, which rules out 1 and 3.
a=2
Only one even number fits in the range.
7.NS.A.2Eliminate Possibilities3STEP 3
Expand each power
Raise every factor inside each parenthesis to its power.
(-x³/y)²=x⁶/y², (y²/x^b)³=y⁶/x^(3b), (-x²/(2y))²=x⁴/(4y²)
(x³)²=x⁶, (y²)³=y⁶, (2y)²=4y².
8.EE.A.1Identify Subproblems4STEP 4
Turn the division into multiplication
Dividing by x⁴/(4y²) means multiplying by 4y²/x⁴. Then collect powers of x and y.
x⁶/y²×y⁶/x^(3b)×4y²/x⁴=4y⁶/(x^(3b-2))
For x: 6-3b-4=2-3b. For y: -2+6+2=6.
8.EE.A.1Convert To Algebra5STEP 5
Match exponents
Compare with 4y^c/x: the powers of y and of x must agree.
c=6, 3b-2=1 → b=1
Same letter, same exponent on both sides.
8.EE.A.1Convert To Algebra6STEP 6
Add
Add the three values.
a+b+c=2+1+6=9
All three are natural numbers, as required.
8.EE.A.1Convert To AlgebraAnswer
9
Put a=2, b=1 back in: x⁶/y²×y⁶/x³×4y²/x⁴=4x⁶⁻³⁻⁴y⁻²⁺⁶⁺²=4y⁶/x, which is the right side with c=6.
Takeaway
Before you chase exponents, check the sign -- it can settle an unknown in one glance.
- Look at the sign first
- Pin down a
- Expand each power
- Turn the division into multiplication
- Match exponents
- Add
Where next?
Level upsuggested
Another one like thissuggested
Step back and solidifysuggested
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