Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**4littlepiggiesmom****Member**- Registered: 2006-01-09
- Posts: 42

y(x^-3/y^-5)^-1 * x(y^3/x^-3)^-3

I'm really lost but have gotten two answers?? 1 isx^6

the other is x^12/y^12

I doubt either are right but I don't knowe what to do about the first letter because I alwasy end up with (an example not the actual factors) x/y^-2/x^1 and don't know what to do next????

Offline

**Tigeree****Member**- Registered: 2005-11-19
- Posts: 13,848

how many toes do u have ?

People don't notice whether it's winter or summer when they're happy.

~ Anton Chekhov

Cheer up, emo kid.

Offline

**4littlepiggiesmom****Member**- Registered: 2006-01-09
- Posts: 42

not enought to count on but 8 sheet f paper later /I have come up with 1/x^5y^13 is that getting closer?

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

First thing is first. A negative power is just the reciprocal of the positive power. The reciprocal is just like flipping a fraction up-side down. Example; x^-2 = 1/x^2.

Next, when you multiply variables with different powers, you keep the variable and add the powers together. Ex; (x^3)(x^2) = x^5.

Last, when you divide variables with different powers, you subtract the power that is in the denominator from the power that is in the numerator. Ex; x^5 / x^2 = x^3.

If you understand all of that, then you should have no problem following this;

Each successive line will further simplify the equation.

y(x^-3/y^-5)^-1 × x(y^3/x^-3)^-3

y[(1/x^3) / (1/y^5)]^-1 × x[y^3 / (1/x^3)]^-3

y(y^5 / x^3)^-1 × x(y^3 x^3)^-3

y(x^3 / y^5) × x(1 / (y^3 x^3)^3)

yx^3 / y^5 × x(1 / (y^9 x^9))

x^3 / y^4 × x / (y^9 x^9)

x^3 / y^4 × 1 / (y^9 x^8)

After multiplication;

x^3 / (y^13 x^8)

1 / (y^13 x^5)

That's your answer. And that is exactly what you posted above. Correct!

Offline