7. Algebra Division of Algebraic Expressions, Exercises of Mathematics

Three sample problems and solutions related to the division of algebraic expressions. The problems involve simplifying and finding the quotient of expressions with variables. The solutions provide step-by-step explanations of the process of simplification and division. useful for students studying algebraic expressions and division in mathematics.

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04 - 22- 23
ALGEBRA: Division of Algebraic Expressions
(Sample Problems & Solutions) - Part 007
1. Find the quotient of the following: ๏ฟฝโˆ’1 ๏ฟฝ2+1 รท
๏ฟฝโˆ’1 ๏ฟฝ+1 .
Solution:
Rewrite the expressions as follows,
๏ฟฝโˆ’1 ๏ฟฝ2+1 รท๏ฟฝโˆ’1 ๏ฟฝ+1 =๏ฟฝโˆ’1 ๏ฟฝ2+1
๏ฟฝโˆ’1 ๏ฟฝ+1
Simplifying the expressions, cancel out ๏ฟฝโˆ’1 ,since theyโ€™re
both present in the numerator and denominator, so we have,
๏ฟฝโˆ’๏ฟฝ ๏ฟฝ2+1
๏ฟฝโˆ’๏ฟฝ ๏ฟฝ+1 =๏ฟฝ๏ฟฝ+๏ฟฝ
๏ฟฝ+๏ฟฝ
Therefore, the quotient of the given problem is ๏ฟฝ๏ฟฝ+๏ฟฝ
๏ฟฝ+๏ฟฝ .
2. Simplify the following: ๏ฟฝ ๏ฟฝโˆ’1 ๏ฟฝ2+1
๏ฟฝ4๏ฟฝโˆ’1 2.
Solution: ๏ฟฝ ๏ฟฝโˆ’1 ๏ฟฝ2+1
๏ฟฝ4๏ฟฝโˆ’1 2
Rewrite the expressions as follows,
๏ฟฝ ๏ฟฝโˆ’1 ๏ฟฝ2+1
๏ฟฝ4๏ฟฝโˆ’1 2
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04 - 22- 23

ALGEBRA: Division of Algebraic Expressions

(Sample Problems & Solutions) - Part 007

  1. Find the quotient of the following: ๐€ โˆ’ 1 ๐€ 2 + 1 รท ๐€ โˆ’ 1 ๐€ + 1 (^). Solution: Rewrite the expressions as follows, ๐€ โˆ’ 1 ๐€^2 + 1 รท ๐€ โˆ’ 1 ๐€ + 1 = ๐€ โˆ’ 1 ๐€^2 + 1 ๐€ โˆ’ 1 ๐€ + 1 Simplifying the expressions, cancel out ๐€ โˆ’ 1 , since theyโ€™re both present in the numerator and denominator, so we have, ๐€ โˆ’ ๐€ ๐€^2 + 1 ๐€ โˆ’ ๐€ ๐€ + 1 = ๐€๐€^ + ๐€ ๐€ + ๐€ Therefore, the quotient of the given problem is ๐€๐€+๐€ ๐€+๐€ .
  2. Simplify the following: ๐€ ๐€โˆ’ 1 ๐€ 2 + ๐€^4 ๐€โˆ’ 1 2

Solution: ๐€ ๐€ โˆ’ 1 ๐€^2 + 1 ๐€ 4 ๐€ โˆ’ 1 2 Rewrite the expressions as follows, ๐€ ๐€ โˆ’ 1 ๐€^2 + 1 ๐€ 4 ๐€ โˆ’ 1 2

๐€ ๐€^ โˆ’^1 ๐€

๐€ โˆ™ ๐€^3 โˆ™ ๐€ โˆ’ 1 โˆ™ ๐€ โˆ’ 1

Cancel out the expressions that has common on both numerator and denominator, as follows, ๐€ โˆ™ ๐€ โˆ’ ๐€ (^) โˆ™ ๐€^2 + 1 ๐€ โˆ™ ๐€^3 โˆ™ ๐€ โˆ’ ๐€ โˆ™ ๐€ โˆ’ 1 ๐€ 2

  • 1 ๐€ 3 ๐€ โˆ’ 1 or ๐€ ๐€
  • ๐€ ๐€ ๐€ โˆ’ ๐€ ๐€ Therefore, simplifying further the given problem, the result is ๐€๐€+๐€ ๐€ ๐€ โˆ’๐€ ๐€.
  1. Find the quotient of the following: ๐€^2 โˆ’ 1 2 ๐€ ๐€+

รท

๐€ ๐€^2 โˆ’ 1 ๐€+

Solution: ๐€ 2 โˆ’ 1 2 ๐€ ๐€ + 1

รท

2 โˆ’ 1 ๐€ + 1 2 Take the reciprocal of the second term, then proceed to multiplication, so we have, ๐€ 2 โˆ’ 1 2 ๐€ ๐€ + 1

รท

2 โˆ’ 1 ๐€ + 1 2

4 ๐€ + 1

รท

๐€ ๐€ + ๐€ ๐€ ๐€ 4 ๐€ + 1

ร—

๐€ ๐€ ๐€ Rewrite the expressions for further simplification as follows, ๐€^4 ๐€ + 1

ร—

3 ๐€ 3 โˆ™ ๐€ ๐€ + 1

ร—

3 Then cancel out some expressions, such as, ๐€๐€^ โˆ™ ๐€ ๐€ + ๐€

ร—

๐€ ๐€ ๐€ + ๐€ Therefore, the quotient of the given problem is ๐€ ๐€ + ๐€.

  1. Find the quotient of the following: ๐€^3 ๐€^2 โˆ’ 1 2 ๐€+

รท

๐€+ 4 ๐€ 4 ๐€^2 โˆ’ 1

Solution: ๐€^3 ๐€^2 โˆ’ 1 2 ๐€ + 1

รท

๐€ + 1 4 ๐€^4

2 โˆ’ 1 Take the reciprocal of the second term, then proceed to multiplication, so we have,

3 ๐€ 2 โˆ’ 1 2 ๐€ + 1

รท

๐€ ๐€ ๐€ ๐€ ๐€ โˆ’ ๐€ ๐€ 3 ๐€ 2 โˆ’ 1 2 ๐€ + 1

ร—

๐€ โˆ’ ๐€ ๐€ + ๐€ ๐€ ๐€ ๐€ Rewrite the expressions for further simplification as follows, ๐€ 3 ๐€ 2 โˆ’ 1 2 ๐€ + 1

ร—

2 โˆ’ 1 ๐€ + 1 4 ๐€ 4 ๐€๐€^ ๐€^2 โˆ’ 1 2 ๐€ + 1

ร—

๐€ + 1 4 ๐€๐€^ โˆ™ ๐€

Then cancel out some expressions, such as, ๐€๐€^ ๐€^2 โˆ’ 1 2 ๐€ + 1

ร—

๐€^2 โˆ’ 1

4 ๐€ ๐€ โˆ™ ๐€ ๐€ 2 โˆ’ 1 2 ๐€ + 1

ร—

2 โˆ’ 1 ๐€ + 1 4 ๐€ ๐€ ๐€ โˆ’ ๐€ ๐€ ๐€ ๐€ + ๐€ ๐€ Therefore, the quotient of the given problem is ๐€ ๐€ โˆ’๐€ ๐€ ๐€ ๐€+๐€ ๐€ . RSM