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
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