# Worksheet on Algebraic Expressions to the Lowest Terms | Simplifying Algebraic Fractions Worksheet

Worksheet on Algebraic Expressions to the Lowest Terms is available here. This worksheet contains the questions and solutions related to the arithmetic operations on the algebraic expressions, simplifying algebraic expressions and reducing them to the lowest terms. Practice various algebraic fraction questions to get knowledge of the concepts. Solve the problems included in Algebraic Expressions to the Lowest Term Worksheets and check answers.

Students can get a detailed solution for every question in the below sections. To express any algebraic expressions in the lowest terms, you need to find the factors of numerator and denominator of the fraction. And cancel the like terms to get the lowest form of the expression. For adding, subtracting two or more algebraic expressions, you should find the least common multiple of denominators and perform the required operation. Check out the various questions at Worksheet on Algebraic Expressions to the Lowest Terms to make a perfect preparation.

1. Reduce the algebraic expressions to their simplest form:

(a) [2(x – 3) / x² – 5x + 6] + [3(x – 1) / x² – 4x + 3] + [5(x – 2) / x² – 3x + 2]

(b) [a / (a² – b²)] – [1 / (a – b)] + 1 / (a + b) + 1/a – 1/b + [(a² – ab + b²) / (ab(a – b))]

(c) (x – y) / xy + (y – z) / yz + (z – x) / zx

Solution:

(a) [2(x – 3) / x² – 5x + 6] + [3(x – 1) / x² – 4x + 3] + [5(x – 2) / x² – 3x + 2]

We observe that the denominators of three fractions are (x² – 5x + 6), (x² – 4x + 3), and (x² – 3x + 2)

The factors of denominators are

= (x² – 3x – 2x + 6), (x² – 3x – x + 3), (x² – 2x – x + 2)

= (x(x – 3) -2(x – 3)), (x(x – 3) – 1(x – 3)), (x(x – 2)-1(x – 2))

= (x-3)(x-2), (x-1)(x-3), (x-1)(x-2)

Therefore, required lowest common multiple of denominators is (x-3)(x-2)(x-1).

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (x-1) ÷ (x-1) in case of [2(x – 3) / x² – 5x + 6], (x – 2) ÷ (x – 2) in case of [3(x – 1) / x² – 4x + 3], (x -3) ÷ (x – 3) in case of [5(x – 2) / x² – 3x + 2].

Therefore, [2(x – 3) / x² – 5x + 6] + [3(x – 1) / x² – 4x + 3] + [5(x – 2) / x² – 3x + 2]

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

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

= [2x²- 8x+ 6 + 3x²- 9x+ 6 + 5x²- 25x+ 30] / (x-3)(x-2)(x-1)

= [10x² – 42x + 42] / (x-3)(x-2)(x-1)

= 2(5x² – 21x +21)/ (x-3)(x-2)(x-1)

(b) [a / (a² – b²)] – [1 / (a – b)] + 1 / (a + b) + 1/a – 1/b + [(a² – ab + b²) / (ab(a – b))]

We observe that the denominators of algebraic fractions are (a² – b²), (a – b), (a + b), a, b, (ab(a – b))

L.C.M of denominators is ab (a² – b²)².

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by ab(a² – b²) ÷ ab (a² – b²) in case of a / (a² – b²), ab(a + b)(a² – b²) ÷ ab(a + b)(a² – b²) in case of 1 / (a – b), ab(a – b)(a² – b²) ÷ ab(a – b)(a² – b²) in case of 1 / (a + b), b(a² – b²)² ÷ b(a² – b²)² in case of 1/a, a(a² – b²)² ÷ a(a² – b²)² in case of 1/b, (a+b) (a² – b²) ÷ (a + b)(a² – b² in case of [(a² – ab + b²) / (ab(a – b)).

Therefore, [a / (a² – b²)] – [1 / (a – b)] + 1 / (a + b) + 1/a – 1/b + [(a² – ab + b²) / (ab(a – b))]

= [a²b(a² – b²) / ab (a² – b²)²] – [ab(a + b)(a² – b²) / ab (a² – b²)²] + [ab(a – b)(a² – b²) / ab (a² – b²)²] + [b(a² – b²)² / ab (a² – b²)²] – [a(a² – b²)²/ ab (a² – b²)²] + [(a+b) (a² – b²)(a² – ab + b²) / ab (a² – b²)²]

= [a²b(a² – b²) – ab(a + b)(a² – b²) + ab(a – b)(a² – b²) + b(a² – b²)² + a(a² – b²)² + (a+b) (a² – b²)(a² – ab + b²)] / ab (a² – b²)²

= [a⁴b-a²b³ – (a⁴b-a²b³+a³b²-ab⁴) + a⁴b-a²b³-a³b²+ab⁴ + ba⁴-2b³a²+b⁵ + a⁵-2a³b²+ab⁴ + a⁵-a³b²+2a²b³-b³a²-b⁵] /ab (a² – b²)²

= [a⁴b-a²b³ – a⁴b+a²b³-a³b²+ab⁴ + a⁴b-a²b³-a³b²+ab⁴ + ba⁴-2b³a²+b⁵ + a⁵-2a³b²+ab⁴ + a⁵-a³b²+2a²b³-b³a²-b⁵] /ab (a² – b²)²

= [2a⁴b – 5a³b² + 3ab⁴ – 2b³a² + 2a⁵] / ab (a² – b²)²

(c) (x – y) / xy + (y – z) / yz + (z – x) / zx

We observe that the denominators of algebraic fractions are xy, yz, zx.

The L.C.M of denominators is xyz.

To make the fractions having a common denominator both the numerator and denominator of these are to be multiplied by z ÷ z in case of (x – y) / xy, x ÷ x in case of (y – z) / yz, y ÷ y in case of (z – x) / zx.

Therefore, (x – y) / xy + (y – z) / yz + (z – x) / zx

= (x – y)z / xyz + (y – z)x / xyz + (z – x)y / xyz

= (xz – yz) /xyz + (xy – xz) / xyz + (yz -xz) / xyz

= (xz – yz + xy – xz + yz – xz) / xyz

= 1 / xyz.

2. Simplify the algebraic expressions by reducing them to their simplest form:

(a) [x² + 2xy + y² / (x – y)] – [x² – 2xy + y²/ (x + y)] – 1 / (x² – y²)

(b) 5a / (a² – 25) + [(a + 2) / (a² + 10a + 25)] – [(a³ + 2a² + a) / (a² – 10a + 25)]

(c) [5x / (2x – 1)] – [(x – 2) / x] + [2x / (x + 2)]

Solution:

(a) [x² + 2xy + y² / (x – y)] – [x² – 2xy + y²/ (x + y)] – 1 / (x² – y²)

We observe that the denominators of three fractions are (x – y), (x + y), and (x² – y²)

Therefore, L.C.M of denominators = (x² – y²)

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (x + y) ÷ (x + y) in case of [x² + 2x + 1 / (x – y)], (x – y) ÷ (x – y) in case of [x² – 2x + 1/ (x + y)].

Therefore, [x² + 2xy + y² / (x – y)] – [x² – 2xy + y²/ (x + y)] – 1 / (x² – y²)

= [(x² + xy + xy + 1) / (x – y)] – [(x² – xy – xy + 1) / (x + y)] – 1 / (x²- y²)

= [(x+y) (x + y) (x + y) / (x – y) (x + y)] – [(x – y) (x – y) (x – y) / (x – y) (x + y)] – 1 / (x² – y²)

= [(x+y) (x + y) (x + y) – (x – y) (x – y) (x – y) – 1] / [(x – y) (x + y)]

= [(x + y)³ – (x – y)³ – 1] / [(x – y) (x + y)]

= [(x³ + y³ + 3x²y + 3xy²) – (x³ – y³ – 3x²y + 3xy²) – 1] / [(x – y) (x + y)]

= [x³ + y³ + 3x²y + 3xy² – x³ + y³ + 3x²y – 3xy² – 1] / [(x – y) (x + y)]

= [2y³ + 6x²y – 1] / [(x – y) (x + y)]

(b) 5a / (a² – 25) + [(a + 2) / (a² + 10a + 25)] – [(a³ + 2a² + a) / (a² – 10a + 25)]

We observe that the denominators of three fractions are (a² – 25), (a² + 10a + 25), and (a² – 10a + 25)

The factors of denominators are

= (a² – 5²), (a² + 5a + 5a + 25), (a² – 5a – 5a + 25)

= (a + 5) (a – 5), (a + 5) (a + 5), (a – 5) (a – 5)

So, L.C.M of denominators is (a – 5)(a – 5)(a + 5)(a + 5)

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (a – 5)(a + 5) ÷ (a – 5)(a + 5) in case of 5a / (a² – 25), (a – 5)(a – 5) ÷ (a – 5)(a – 5) in case of [(a + 2) / (a² + 10a + 25)], (a + 5)(a + 5) ÷(a + 5)(a + 5) in case of [(a³ + 2a² + a) / (a² – 10a + 25)].

Therefore, 5a / (a² – 25) + [(a + 2) / (a² + 10a + 25)] – [(a³ + 2a² + a) / (a² – 10a + 25)]

= 5a (a – 5)(a + 5)/ (a – 5)(a – 5)(a + 5)(a + 5) + [(a + 2)(a – 5)(a – 5) / (a – 5)(a – 5)(a + 5)(a + 5)] – [(a³ + 2a² + a) (a + 5)(a + 5) / (a – 5)(a – 5)(a + 5)(a + 5)]

= [5a (a – 5)(a + 5) + (a + 2)(a – 5)(a – 5) – (a³ + 2a² + a) (a + 5)(a + 5)] / (a – 5)(a – 5)(a + 5)(a + 5)

= [5a (a – 5)(a + 5) + (a + 2)(a – 5)(a – 5) – a(a² + 2a + 1) (a + 5)(a + 5)] / (a – 5)(a – 5)(a + 5)(a + 5)

= [5a (a – 5)(a + 5) + (a + 2)(a – 5)(a – 5) – a (a + 1) (a + 1) (a + 5)(a + 5)] / (a – 5)(a – 5)(a + 5)(a + 5)

= [5a³ – 125a + a³ -8a² + 5a + 50 – (a⁵ +12a⁴ +46a³ +60a² + 25a)] / (a – 5)(a – 5)(a + 5)(a + 5)

= [6a³ – 8a² – 120a + 50 – a⁵ -12a⁴ -46a³ -60a² – 25a] / (a – 5)(a – 5)(a + 5)(a + 5)

= [- a⁵ -12a – 40a³ – 68a² – 140a + 50] / (a – 5)(a – 5)(a + 5)(a + 5)

(c) [5x / (2x – 1)] – [(x – 2) / x] + [2x / (x + 2)]

We observe that the denominators of three fractions are (2x – 1), x, and (x + 2)

The L.C.M of denominators is x(2x – 1) (x + 2).

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by x(x + 2) ÷ x(x + 2) in case of 5x / (2x – 1), (x + 2) (2x – 1) ÷ (x + 2) (2x – 1) in case of [(x – 2) / x], x(2x – 1) ÷x(2x – 1) in case of [2x / (x + 2)].

Therefore, [5x / (2x – 1)] – [(x – 2) / x] + [2x / (x + 2)]

= [5x²(x + 2) / x(2x – 1) (x + 2)] – [(x + 2) (2x – 1)(x – 2)/ x(2x – 1) (x + 2)] + [2x²(2x – 1)/ x(2x – 1) (x + 2)]

= [5x²(x + 2) – (x + 2) (2x – 1)(x – 2) + 2x²(2x – 1)] / x(2x – 1) (x + 2)

= [5x³ + 10x² – 2x³ – x² – 8x + 4 + 4x³ – 2x²] / x(2x – 1) (x + 2)

= [7x³ + 7x² – 8x + 4] / x(2x – 1) (x + 2)

3. Reduce by multiplying and dividing the algebraic fractions to its lowest term:

(a) [a(a + b) / (a – b)] x [(a – b) / b(a + b)] x a/b

(b) [(x² – 2x) / x² + 3x – 10)] ÷ [(x² + 4x – 21) / (x² + 2x – 15)]

(c) [(a² – 15a + 4)/(a² – 7a + 10)] x [(a² – a – 2) / (a² + 2a – 3)] ÷ [(a² – 5a + 4) / (a² + 8a + 15)]

Solution:

(a) [a(a + b) / (a – b)] x [(a – b) / b(a + b)] x a/b

Multiply the numerators together and denominators together.

= [a(a + b) (a – b) a] / [(a – b) (a + b) b b]

= [a² (a + b) (a – b)] / [b² (a + b) (a – b)

Cancel the common terms ( a+ b), (a – b) in both numerator and denominators.

= a² / b², which is the reduced form of the algebraic expression.

(b) [(x² – 2x) / (x² + 3x – 10)] ÷ [(x² + 4x – 21) / (x² + 2x – 15)]

Find the factors of the numerator and denominator of the algebraic expressions.

= [x(x – 2) / (x² + 5x – 2x – 10)] ÷ [(x² + 7x – 3x – 2) / (x² +5x – 3x – 15)]

= [x(x – 2) / (x(x + 5) – 2(x + 5)] ÷ [(x(x + 7) – 3 (x + 7)) / (x (x + 5) – 3 (x + 5))]

= [x(x – 2)/ (x – 2) (x + 5)] ÷ [(x + 7) (x – 3) / (x + 5) (x – 3)]

Cancel the common terms in the numerator and denominators.

= [x / (x + 5)] ÷ [(x + 7) / (x + 5)]

Reverse the second fraction and multiply with the first one.

= [x / (x + 5)] x [(x + 5) / (x + 7)]

= [x(x + 5) / (x + 5) (x + 7)]

Cancel the common term (x + 5) in both numerator and denominator.

= x / (x + 7), which is the lowest term of the algebraic expression.

(c) [(a² – 15a + 4)/(a² – 7a + 10)] x [(a² – a – 2) / (a² + 2a – 3)] ÷ [(a² – 5a + 4) / (a² + 8a + 15)]

Find the factors of the numerator and denominator of the algebraic expressions.

= [(a² – 15a + 4) / (a² – 5a – 2a + 10)] x [(a² – 2a + a – 2) / (a² + 3a – a  – 3)] ÷ [(a² – 4a – a + 4) / (a² + 5a + 3a + 15)]

= [(a² – 15a + 4) / (a(a – 5) – 2(a – 5))] x [(a(a – 2) + 1(a – 2) / (a(a+3) – 1(a +3)] ÷ [(a(a – 4) -1(a – 4) / (a(a + 5) + 3(a + 5)]

= [(a² – 15a + 4) / (a – 5)(a – 2)] x [(a – 2)(a + 1) / (a – 1) (a + 3)] ÷ [(a -1) (a -4) / (a + 3) (a + 5)]

= [(a² – 15a + 4) (a – 2)(a + 1)/ (a – 5)(a – 2)(a – 1) (a + 3)] ÷ [(a -1) (a -4) / (a + 3) (a + 5)]

Cancel the common terms in the numerator and denominator.

= [(a² – 15a + 4) (a + 1) / (a – 5) (a + 3)] x [(a + 3) (a + 5) / (a -1) (a -4)]

= [(a² – 15a + 4) (a + 1) (a + 3) (a + 5) / (a – 5) (a + 3)(a -1) (a -4)]

= [(a² – 15a + 4) (a + 1) (a + 5) / (a – 5) (a – 4)]