Here we will learn how to simplify the problems on algebraic fractions to its lowest term.
1. Reduce the algebraic fractions to their lowest terms: \(\frac{x^{2} - y^{2}}{x^{3} - x^{2}y}\)
Solution:
\(\frac{x^{2} - y^{2}}{x^{3} - x^{2}y}\)
Factorizing the numerator and denominator separately and cancelling the common factors we get,
= \(\frac{(x + y) (x - y)}{x^{2} (x - y)} \)
= \(\frac{x + y}{x^{2}}\)
2. Reduce to lowest terms \(\frac{x^{2} + x - 6}{x^{2} - 4}\)
Solution:
\(\frac{x^{2} + x - 6}{x^{2} - 4}\)
Step 1: Factorize the numerator x\(^{2}\) + x – 6
= x\(^{2}\) + 3x – 2x – 6
= x(x + 3) – 2(x + 3)
= (x + 3) (x – 2)
Step 2: Factorize the denominator: x\(^{2}\) – 4
= x\(^{2}\) – 2\(^{2}\)
= (x + 2) (x – 2)
Step 3: From steps 1 and 2: \(\frac{x^{2} + x - 6}{x^{2} - 4}\)
= \(\frac{x^{2} + x - 6}{x^{2} - 2^{2}}\)
= \(\frac{(x + 3) (x - 2)}{(x + 2) (x - 2)}\)
= \(\frac{(x + 3)}{(x + 2)}\)
3. Simplify the algebraic fractions \(\frac{36x^{2} - 4}{9x^{2} + 6x + 1}\)
Solution:
\(\frac{36x^{2} - 4}{9x^{2} + 6x + 1}\)
Step 1: Factorize the numerator: 36x\(^{2}\) – 4
= 4(9x\(^{2}\) – 1)
= 4[(3x)\(^{2}\) – (1)\(^{2}\)]
= 4(3x + 1) (3x – 1)
Step 2: Factorize the denominator: 9x\(^{2}\) + 6x + 1
= 9x\(^{2}\) + 3x + 3x + 1
= 3x(3x + 1) + 1(3x + 1)
= (3x + 1) (3x + 1)
Step 3: Simplification of the given expression after factorizing the numerator and the denominator:
\(\frac{36x^{2} - 4}{9x^{2} + 6x + 1}\)
= \(\frac{4(3x + 1)(3x - 1)}{(3x + 1)(3x + 1)}\)
= \(\frac{4(3x - 1)}{(3x + 1)}\)
4. Reduce and simplify: \(\frac{8x^{3}y^{2}z}{2xy^{3}} of \left ( \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \div \frac{7xy^{2}}{35x^{2}yz^{3}}\right )\)
Solution:
\(\frac{8x^{3}y^{2}z}{2xy^{3}} of \left ( \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \div \frac{7xy^{2}}{35x^{2}yz^{3}}\right )\)
= \(\frac{8x^{3}y^{2}z}{2xy^{3}} of \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \times \frac{35x^{2}yz^{3}}{7xy^{2}}\)
= \(\frac{4x^{3}y^{2}z}{xy^{3}} \left ( \frac{x^{5}y^{2}z^{2}}{xy^{3}z} \times \frac{x^{2}yz^{3}}{xy^{2}} \right )\)
= 4x\(^{10 - 3}\) ∙ y\(^{-3}\) ∙ z\(^{5}\)
= \(\frac{4x^{7}\cdot z^{5}}{y^{3}}\)
5. Simplify: \(\frac{2x^{2} - 3x - 2}{x^{2} + x - 2} \div \frac{2x^{2} + 3x + 1}{3x^{2} + 3x - 6}\)
Solution:
\(\frac{2x^{2} - 3x - 2}{x^{2} + x - 2} \div \frac{2x^{2} + 3x + 1}{3x^{2} + 3x - 6}\)
Step 1: First factorize each of the polynomials separately:
2x\(^{2}\) – 3x – 2 = 2x\(^{2}\) – 4x + x – 2
= 2x(x – 2) + 1 (x – 2)
= (x – 2) (2x + 1)
x\(^{2}\) + x – 2 = x\(^{2}\) + 2x - x – 2
= x(x + 2) - 1 (x + 2)
= (x + 2) (x - 1)
2x\(^{2}\) + 3x + 1 = 2x\(^{2}\) + 2x + x + 1
= 2x(x + 1) + 1 (x + 1)
= (x + 1) (2x + 1)
3x\(^{2}\) + 3x – 6 = 3[x\(^{2}\) + x – 2]
= 3[x\(^{2}\) + 2x - x – 2]
= 3[x(x + 2) – 1(x + 2)]
= 3[(x + 2) (x - 1)]
= 3[(x + 2) (x - 1)]
= 3(x + 2) (x - 1)
Step 2: Simplify the given expressions by substituting with their factors
\(\frac{2x^{2} - 3x - 2}{x^{2} + x - 2} \div \frac{2x^{2} + 3x + 1}{3x^{2} + 3x - 6}\)
= \(\frac{2x^{2} - 3x - 2}{x^{2} + x - 2} \times \frac{3x^{2} + 3x - 6}{2x^{2} + 3x + 1}\)
= \(\frac{(x - 2) (2x + 1)}{(x + 2) (x - 1)}\times\frac{3(x + 2) (x - 1)}{(x + 1) (2x + 1)}\)
= \(\frac{3(x - 2)}{(x + 1)}\)
8th Grade Math Practice
From Problems on Algebraic Fractions to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Nov 25, 24 01:18 AM
Nov 25, 24 01:09 AM
Nov 25, 24 12:48 AM
Nov 25, 24 12:17 AM
Nov 24, 24 11:01 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.