Chapter 4.7, Problem 11PS

Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

Chapter
Section

Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

For Problem 1-30, solve each equation. (Objective 1) 2 n − 2 − n n + 5 = 10 n + 15 n 2 + 3 n − 10

To determine

To solve:

The equation 2n2nn+5=10n+15n2+3n10.

Explanation

Approach:

Rational expression, a quotient obtained by division of two polynomials in form of p(x,y)q(x,y) where p(x,y) and q(x,y) are polynomials in such a way that the variables x, y do not assume values such that q(x,y)=0

If ab and cd are rational, then addition or subtraction of rational formed by making the denominator equal using LCD and perform the operation on the numerators as follows:

And,

Same algebra is for rational expression.

Calculation:

The given rational expression is 2nâˆ’2âˆ’nn+5=10n+15n2+3nâˆ’10

Factorize each term of the equation,

2nâˆ’2âˆ’nn+5=10n+15(n+5)(nâˆ’2)

Restriction on the equation is that nâ‰ 2â€‰andâ€‰nâ‰ âˆ’5,

The lowest common denominator of the given equation is (nâˆ’2)(n+5)

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