Algebraic Fractions — Solutions

Full worked solutions, with method and final answer for every part.

Question 1

(a)Use a common denominator of x(x+1): 2/x becomes 2(x+1)/[x(x+1)], and 3/(x+1) becomes 3x/[x(x+1)], then add the numerators.

[2(x+1) + 3x] / [x(x+1)] = (5x + 2) / [x(x+1)]

Question 2

(a)Factorise the numerator as a difference of two squares: x² − 9 = (x−3)(x+3), then cancel the common factor (x+3).

x − 3

Question 3

(a)Common denominator x(x+2): 3/x becomes 3(x+2)/[x(x+2)], 2/(x+2) becomes 2x/[x(x+2)].

[3(x+2) − 2x] / [x(x+2)] = (x + 6) / [x(x+2)]

Question 4

(a)Factorise the numerator (difference of squares) and cancel the common factor.

(x−5)(x+5)/(x−5) = x + 5

Question 5

(a)Cancel the common factor of 2x from top and bottom.

2/x

Question 6

(a)Cancel the common factor (x+2) from top and bottom.

1/(x−3), undefined when x = −2 or x = 3

Question 7

(a)Multiply numerators and denominators, then simplify: (5×4x)/(2x×15) = 20x/30x.

2/3

Question 8

(a)Common denominator 2x: 1/x = 2/(2x).

2/(2x) + 1/(2x) = 3/(2x)

Question 9

(a)Common denominator (x+1)(x−1): combine numerators 2(x−1)+1(x+1).

(3x − 1)/(x² − 1)

Question 10

(a)Dividing by a negative power means adding the positive index.

x⁶

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