Unit 8: Algebraic Fractions
Exercise 8.1 — WordPress-ready solutions reconstructed from the supplied 18-page handwritten notes.
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Useful Factorization Identities Used in the Solutions
Exercise 8.1
1. Change into the Simplest Form
(a) (n² − 5n)/(n² − 25)
= n(n−5) / [n² − 5²]
= n(n−5) / [(n+5)(n−5)]
∴ n/(n+5)
(b) (m² − b²)/(m+b)²
= (m+b)(m−b) / [(m+b)(m+b)]
∴ (m−b)/(m+b)
(c) (n² − 5n + 6)/(n² − 7n + 12)
= [n² − 3n − 2n + 6] / [n² − 4n − 3n + 12]
= [(n−2)(n−3)] / [(n−3)(n−4)]
∴ (n−2)/(n−4)
2. Simplify — Parts (a) to (e)
(a) a/(a−b) + b/(b−a)
= a/(a−b) − b/(a−b)
= (a−b)/(a−b)
∴ 1
(b) 1/(b−c) − (b+c)/(b²−c²)
b²−c² = (b−c)(b+c)
= 1/(b−c) − (b+c)/[(b−c)(b+c)]
∴ 0
(c) 1/(m−n) + 1/(m+n)
= [(m+n)+(m−n)] / (m²−n²)
∴ 2m/(m²−n²)
(d) (m+n)/(m−n) + (m−n)/(m+n)
= [(m+n)² + (m−n)²] / (m²−n²)
= [2m² + 2n²] / (m²−n²)
∴ 2(m²+n²)/(m²−n²)
(e) 1/(m+n) + n/(m²−n²)
= [(m−n)+n] / [(m−n)(m+n)]
∴ m/(m²−n²)
2. Simplify — Parts (f) to (j)
(f) 3/(m²−4) + 1/(m−2)²
= 3/[(m−2)(m+2)] + 1/(m−2)²
= [3(m−2)+(m+2)] / [(m−2)²(m+2)]
= (4m−4) / [(m−2)²(m+2)]
∴ 4(m−1)/[(m−2)²(m+2)]
(g) (a³+b³)/(a²−ab+b²) + (a³−b³)/(a²+ab+b²)
= (a+b) + (a−b)
∴ 2a
(h) (4m²+25y²)/(4m²−25y²) − (2m−5y)/(2m+5y)
4m²−25y² = (2m+5y)(2m−5y)
= [(2m+5y)² − 20my − (2m−5y)²] / [(2m+5y)(2m−5y)]
∴ 20my/(4m²−25y²)
(i) 4m³/(m⁴+a⁴) − 8m⁷/(m⁸−a⁸)
m⁸−a⁸ = (m⁴−a⁴)(m⁴+a⁴)
= [4m³(m⁴−a⁴) − 8m⁷] / [(m⁴+a⁴)(m⁴−a⁴)]
= −4m³(m⁴+a⁴) / [(m⁴+a⁴)(m⁴−a⁴)]
∴ −4m³/(m⁴−a⁴)
(j) m/(m−y) − m/(m+y) + 2my/(m²+y²)
The first two terms give 2my/(m²−y²).
Combining with 2my/(m²+y²):
∴ 4m³y/(m⁴−y⁴)
2. Simplify — Parts (k) to (o)
(k) 3/(a+3) + 4/(a−3) + 9a/[2(9−a²)]
The first two fractions combine to (7a+3)/(a²−9).
Since 9a/[2(9−a²)] = −9a/[2(a²−9)],
∴ (5a+6)/[2(a²−9)]
(l) 1/(m+2y) − 1/(m−2y) + 2m/(4y²−m²)
4y²−m² = −(m−2y)(m+2y)
Combining all three fractions:
∴ −2/(m−2y)
(m) a/[(a−b)(a−c)] + b/[(b−a)(b−c)] + c/[(c−b)(c−a)]
Taking the common denominator and simplifying the numerator gives 0.
∴ 0
(n) (y−z)/[m²−(y−z)²] + (z−m)/[y²−(z−m)²] + (m−y)/[z²−(m−y)²]
Each denominator is factorized as a difference of squares.
After taking the common denominator, all numerator terms cancel.
∴ 0
(o) Cyclic algebraic-fraction expression
After factorizing the differences of squares, the three fractions have the same denominator (m+a+b).
The numerators add to m+a+b.
∴ 1
2. Simplify — Parts (p) to (u)
(p) 1/(p²+7p+12) + 2/(p²+5p+6) − 3/(p²+6p+8)
p²+7p+12 = (p+3)(p+4)
p²+5p+6 = (p+2)(p+3)
p²+6p+8 = (p+2)(p+4)
Combining:
∴ 1/[(p+4)(p+3)(p+2)]
(q) (m+3)/(m²+3m+9) + (m−3)/(m²−3m+9) − 54/(m⁴+9m²+81)
(m²+3m+9)(m²−3m+9) = m⁴+9m²+81
The first two terms combine to 2m³/(m⁴+9m²+81).
= 2(m³−27)/[(m²+3m+9)(m²−3m+9)]
∴ 2(m−3)/(m²−3m+9)
(r) 1/(m²−5m+6) + 2/(4m−m²−3) − 3/(m²−3m+2)
m²−5m+6 = (m−2)(m−3)
4m−m²−3 = −(m−1)(m−3)
m²−3m+2 = (m−1)(m−2)
After simplification:
∴ −4/(m²−3m+2)
(s) (b+2)/(1+b+b²) − (b−2)/(1−b+b²) − 2b²/(1+b²+b⁴)
Using (1+b+b²)(1−b+b²) = 1+b²+b⁴, the terms reduce to:
∴ 4/(1+b²+b⁴)
(t) 1/(1−b+b²) − 1/(1+b+b²) − 2b/(1−b²+b⁴)
The first two fractions combine to 2b/(1+b²+b⁴).
Combining with the third fraction:
∴ −4b³/(1+b⁴+b⁸)
(u) (a+c)/(a²+ac+c²) + (a−c)/(a²−ac+c²) + 2c³/(a⁴+a²c²+c⁴)
Using (a²+ac+c²)(a²−ac+c²) = a⁴+a²c²+c⁴,
the expression simplifies to:
∴ 2(a+c)/(a²+ac+c²)
3. Simplify
(a) 1/[4(1−√n)] − 1/[4(1+√n)] + 2√n/[4(1−n)]
The first two fractions are rationalized through the common factor (1−√n)(1+√n)=1−n.
They give 2√n/[4(1−n)].
Adding the third fraction:
∴ √n/(1−n)
(b) 1/[8(1−√n)] − 1/[8(1+√n)] + 2√n/[8(1−n)]
Following the same method:
∴ √n/[2(1−n)]
(c) 1/(a+1)² + 1/(a−1)² − 2/(a²−1)
a²−1 = (a−1)(a+1)
Taking the common denominator (a²−1)²:
Numerator = (a−1)² + (a+1)² − 2(a+1)(a−1)
= 4
∴ 4/(a²−1)²
4. Find the Value of a
Since 2n²+5n+2 = (2n+1)(n+2),
[a(n+2) + (2n+1)] / [(2n+1)(n+2)] = (4n+5)/(2n²+5n+2)
(a+2)n + (2a+1) = 4n + 5
Comparing coefficients:
a + 2 = 4
∴ a = 2
5. Find the Values of a and b
6n²−n−12 = (2n−3)(3n+4)
a(3n+4) + b(2n−3) = n+7
(3a+2b)n + (4a−3b) = n+7
Comparing coefficients:
3a + 2b = 1 … (i)
4a − 3b = 7 … (ii)
Multiplying (i) by 3 and (ii) by 2, then adding:
17a = 17
∴ a = 1
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