Q. \[ \frac{2 x^4 y^{-4} z^{-3}}{3 x^2 y^{-3} z^4} \]

Q: How do I simplify \frac{2x^{4}y^{-4}z^{-3}}{3x^{2}y^{-3}z^{4}} ?

Use the quotient rule \frac{a^m}{a^n}=a^{m-n}: coefficient \frac{2}{3}, x^{4-2}=x^2, y^{-4-(-3)}=y^{-1}, z^{-3-4}=z^{-7}. Result: \frac{2}{3}x^2y^{-1}z^{-7} or \frac{2x^2}{3yz^7}.

Q: How do I remove negative exponents?

Move factors between numerator and denominator: y^{-1} = 1/y, z^{-7} = 1/z^7. So \frac{2}{3}x^2y^{-1}z^{-7} = \frac{2x^2}{3yz^7}.

Q: Why do we subtract exponents when dividing like bases?

Because a^{m}/a^{n}=a^{m-n}, derived from canceling n factors of a. Subtraction gives the remaining power after division.

Q: Can I simplify the numeric coefficient 2/3 further?

No, 2/3 is already in lowest terms; nothing cancels between 2 and 3.

Q: Are there any domain restrictions for x,y,z?

Yes. Any variable in a denominator must be nonzero. From \frac{2x^2}{3yz^7} we require y \neq 0 and z \neq 0. x may be 0.

Q: Does the result change if x,y,z are negative?

Does the result change if x,y,z are negative?

Q: How can I check my simplification is correct?

Plug simple nonzero values (e.g., x=2, y=1, z=1) into original and simplified expressions; both evaluate to the same number.

Answer

\( \frac{2x^{4}y^{-4}z^{-3}}{3x^{2}y^{-3}z^{4}} = \frac{2}{3}x^{4-2}y^{-4-(-3)}z^{-3-4} = \frac{2}{3}x^{2}y^{-1}z^{-7} = \frac{2x^{2}}{3yz^{7}} \)

Detailed Explanation

Problem:

\[ \frac{2x^{4}y^{-4}z^{-3}}{3x^{2}y^{-3}z^{4}} \]

Separate the numerical coefficient from each variable factor.

Write the expression as a product of quotients:

\[ \frac{2}{3} \cdot \frac{x^{4}}{x^{2}} \cdot \frac{y^{-4}}{y^{-3}} \cdot \frac{z^{-3}}{z^{4}} \]
Apply the quotient rule for exponents to each variable (\( \frac{a^{m}}{a^{n}} = a^{m-n} \)).

Compute exponents separately:
- For \( x \): the exponent is 4 minus 2, so \( x^{4-2} = x^{2} \).
- For \( y \): the exponent is -4 minus (-3), so \( y^{-4-(-3)} = y^{-1} \).
- For \( z \): the exponent is -3 minus 4, so \( z^{-3-4} = z^{-7} \).
So the expression becomes

\[ \frac{2}{3} \, x^{2} \, y^{-1} \, z^{-7} \]
Remove negative exponents by moving factors to the denominator.

Recall that \( a^{-k} = \frac{1}{a^{k}} \).
- \( y^{-1} = \frac{1}{y} \)
- \( z^{-7} = \frac{1}{z^{7}} \)
Thus

\[ \frac{2}{3} \, x^{2} \, y^{-1} \, z^{-7} = \frac{2x^{2}}{3 \, y \, z^{7}} \]
Final simplified form.
\[ \frac{2x^{2}}{3y z^{7}} \]

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FAQs

How do I simplify \( \frac{2x^{4}y^{-4}z^{-3}}{3x^{2}y^{-3}z^{4}} \)?

Use the quotient rule \(\frac{a^m}{a^n}=a^{m-n}\): coefficient \(\frac{2}{3}\), \(x^{4-2}=x^2\), \(y^{-4-(-3)}=y^{-1}\), \(z^{-3-4}=z^{-7}\). Result: \(\frac{2}{3}x^2y^{-1}z^{-7}\) or \(\frac{2x^2}{3yz^7}\).

How do I remove negative exponents?

Move factors between numerator and denominator: \(y^{-1} = 1/y\), \(z^{-7} = 1/z^7\). So \(\frac{2}{3}x^2y^{-1}z^{-7} = \frac{2x^2}{3yz^7}\).

Why do we subtract exponents when dividing like bases?

Because \(a^{m}/a^{n}=a^{m-n}\), derived from canceling \(n\) factors of \(a\). Subtraction gives the remaining power after division.

Can I simplify the numeric coefficient \(2/3\) further?

No, \(2/3\) is already in lowest terms; nothing cancels between 2 and 3.

Are there any domain restrictions for \(x,y,z\)?

Yes. Any variable in a denominator must be nonzero. From \(\frac{2x^2}{3yz^7}\) we require \(y \neq 0\) and \(z \neq 0\). \(x\) may be 0.

Does the result change if \(x,y,z\) are negative?

How can I check my simplification is correct?

Plug simple nonzero values (e.g., \(x=2, y=1, z=1\)) into original and simplified expressions; both evaluate to the same number.

How would I show the steps clearly?

Write numerator and denominator separated, cancel common factors (coefficients and like bases), apply \(m-n\) for each base, then rewrite to remove negatives. Example steps yield \(\frac{2x^2}{3yz^7}\).

Simplify the expression step by step.
Apply exponent rules to simplify.

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