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# 6x^7*y^5/3x^-1

Please solve! Six 'x' to the seventh power times 'y' to the fifth power over three 'x' to the power of negative one.

### 1 Answer by Expert Tutors

Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
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(6x7y5)/(3x-1)

First, notice that the 6 (in the numerator) and the 3 (in the denominator are coefficients of the product of the terms in the numerator and the denominator, respectively. This means that these constant terms are being divided by one another and can be simplified to 2, since 6/3 = 2. In other words, if you factor out a 3 from both the numerator and the denominator, then these terms cancel each other out:

3(2x7y5)/3(x-1) = (3/3)(2x7y5/x-1) = 2x7y5/x-1

Note that since the term in the denominator is to the power of a -1, then we can make this negative exponent positive by taking its reciprocal. That is, if we move the 'x-1' in the denominator to the numerator then the exponent becomes a positive 1 or the term becomes 'x1':

2x7y5/x-1 = 2x7y5x1

When multiplying exponential terms with like bases, then we keep the same base and add their exponents:

2x7y5x1 = 2x7x1y5 = 2x7+1y5 = 2x8y5