Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
4. Rational Expressions
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Exercise 21 Page 531

We want to divide the given rational expressions. 3x^3/5y^2÷ 6y^(-3)/5x^(-5) To divide the expressions, let's recall the Negative Exponent Property. x^(- n)=1/x^nThis property can be rewritten for cases where we have a negative exponent in the denominator. 1/x^(- n)=x^n Let's use the above to start simplifying the expressions. We will also split the numerators and denominators of both expressions into factors to cancel out any common factors.
3x^3/5y^2÷ 6y^(-3)/5x^(-5)
3x^3/5y^2÷ 6y^(-3)/5* 1/x^(-5)
3x^3/5y^2÷ 6/5* y^(-3)* 1/x^(-5)
3x^3/5y^2÷ 6/5* 1/y^3* 1/x^(-5)

1/x^(- n)=x^n

3x^3/5y^2÷ 6/5* 1/y^3* x^5
3x^3/5y^2÷ 6/5y^3* x^5
3x^3/5y^2÷ 6x^5/5y^3
3x^3/5y^2÷ 6x^(3+2)/5y^(2+1)
3x^3/5y^2÷ 6x^3* x^2/5y^2* y^1
3x^3/5y^2÷ 6x^3* x^2/5y^2* y
3x^3/5y^2÷ 2* 3x^3* x^2/5y^2* y
To divide the expressions, we will multiply the first expression by the reciprocal of the second expression. 3x^3/5y^2÷ 2* 3x^3* x^2/5y^2* y ⇕ 3x^3/5y^2* 5y^2* y/2* 3x^3* x^2 Finally, we can cancel out any common factors.
3x^3/5y^2* 5y^2* y/2* 3x^3* x^2
3x^3/5y^2* 5y^2* y/2* 3x^3* x^2
y/2* x^2
y/2x^2
To identify the restrictions on the variables, we need to find any values of x and y that would cause the denominator of the simplified expression, or any other denominator used, to be 0.
Denominator Restrictions on the denominator Restrictions on the variables
5y^2 5y^2≠ 0 y≠ 0
5x^(-5) x^5≠ 0 x≠ 0
5y^2* y y^2≠ 0 and y≠0 y≠ 0 and y≠ 0
2* 3x^3* x^2 3x^3≠ 0 and x^2≠ 0 x≠ 0 and x≠ 0
2x^2 2x^2≠ 0 x≠ 0

We found two restrictions on the variables. y≠ 0, x≠ 0