Algebraic Fractions: Simplify, Add & Subtract (Grade 10)

∑ DANEMATHICS

FREE CAPS MATHS RESOURCES

Grade 10 · Algebraic Expressions

MARKING GUIDELINE

Marks

Duration

55 minutes

Questions

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Class:

Date:

Mark

/ 35

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
x² − 4x + 2 simplifies to …
(1)
A)x − 2✓
B)x + 2
C)x² − 2
D)−2
Answer: A — x² − 4 = (x + 2)(x − 2), so the (x + 2) cancels.
- B — cancelled the wrong factor
- C — cancelled the 4 against the 2 without factorising first
- D — cancelled x² against x, which cannot be done across a sum
1.2
x3 + x4 = …
(1)
A)7x12✓
B)2x7
C)x12
D)2x12
Answer: A — The lowest common denominator is 12, so it is 4x/12 + 3x/12 = 7x/12.
- B — added the numerators AND the denominators
- C — used the common denominator but forgot to rewrite the numerators
- D — added the numerators without first writing each over 12
1.3
3x + 6x² − 4 simplifies to …
(1)
A)3x − 2✓
B)3x + 2
C)3xx − 2
D)3(x + 2)x − 2
Answer: A — 3x + 6 = 3(x + 2) and x² − 4 = (x + 2)(x − 2), so the (x + 2) cancels.
- B — cancelled (x − 2) instead of (x + 2)
- C — cancelled the 6 against the 4 and left the x in the numerator
- D — cancelled the denominator's (x + 2) but not the numerator's
1.4
x2 ÷ x6 = …
(1)
A)3✓
B)13
C)x²12
D)x3
Answer: A — Dividing means multiplying by the reciprocal: (x/2) × (6/x) = 6x/2x = 3.
- B — inverted the FIRST fraction instead of the second
- C — multiplied straight across without inverting the second fraction
- D — cancelled the x and then wrote it in again
1.5
2x − 1x + 1 = …
(1)
A)x + 2x(x + 1)✓
B)1x(x + 1)
C)12x + 1
D)x + 22x + 1
Answer: A — Over the LCD x(x + 1) this is (2x + 2 − x)/(x(x + 1)) = (x + 2)/(x(x + 1)).
- B — subtracted the numerators without first multiplying each by its missing factor
- C — subtracted the numerators AND the denominators
- D — found the right numerator but subtracted the denominators
1.6
For which value of x is x + 3x − 5 UNDEFINED?
(1)
A)x = 5✓
B)x = −3
C)x = −5
D)x = 3
Answer: A — A fraction is undefined only when its DENOMINATOR is zero, and x − 5 = 0 at x = 5.
- B — that makes the NUMERATOR zero, which gives 0, not undefined
- C — x − 5 = 0 gives x = +5, not −5
- D — changed the sign of the numerator's 3 instead of solving the denominator
1.7
x² + 5x + 6x² − 9 simplifies to …
(1)
A)x + 2x − 3✓
B)x + 2x + 3
C)x + 3x − 3
D)5x + 6−9
Answer: A — (x + 2)(x + 3) over (x + 3)(x − 3) — the (x + 3) cancels.
- B — cancelled (x − 3) instead of (x + 3)
- C — cancelled (x + 2) instead of (x + 3)
- D — cancelled the x² terms, which cannot be done across a sum
1.8
x + 1x − 2 × x − 2x + 3 = …
(1)
A)x + 1x + 3✓
B)x − 2x + 3
C)x + 1x − 2
D)(x + 1)(x − 2)(x − 2)(x + 3)
Answer: A — The (x − 2) appears in a numerator and in a denominator, so it cancels.
- B — cancelled (x + 1) instead of (x − 2)
- C — cancelled (x + 3) instead of (x − 2)
- D — a correct product, but the common (x − 2) has not been cancelled
1.9
x³ − 8x − 2 simplifies to …
(1)
A)x² + 2x + 4✓
B)x² − 2x + 4
C)x² − 4
D)x² + 4
Answer: A — x³ − 8 = (x − 2)(x² + 2x + 4), so the (x − 2) cancels.
- B — the middle sign of the trinomial is the OPPOSITE of the binomial's
- C — treated x³ − 8 as a difference of squares
- D — left out the middle term 2x
1.10
2x − 69 − x² simplifies to …
(1)
A)−23 + x✓
B)23 + x
C)23 − x
D)−23 − x
Answer: A — 2x − 6 = −2(3 − x) and 9 − x² = (3 − x)(3 + x), so the (3 − x) cancels and the minus stays.
- B — lost the minus that comes from reversing x − 3 into 3 − x
- C — cancelled (3 + x) instead of (3 − x), and lost the minus too
- D — cancelled (3 + x) instead of (3 − x)

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[11 MARKS]

Simplify the following FULLY.

2.1
x² − 16x + 4
(3)
Factorise the numerator: (x + 4)(x − 4)x + 4 (2)
= x − 4 (1)
2.2
x² − 1x ÷ x + 13x
(4)
Multiply by the reciprocal (1)
= (x + 1)(x − 1) × 3xx(x + 1) (2)
= 3(x − 1) (1)
2.3
2x + 1 + 3x − 1
(4)
LCD = (x + 1)(x − 1) (1)
= 2(x − 1) + 3(x + 1)x² − 1 (2)
= 5x + 1x² − 1 (1)

Question 3

[14 MARKS]

Simplify the following FULLY, and answer the question that follows.

3.1
x − 2x³ − 8
(4)
The denominator is a difference of two cubes (1)
= x − 2(x − 2)(x² + 2x + 4) (2)
= 1x² + 2x + 4 (1)
3.2
x² + 5x + 6x² − 4
(4)
= (x + 2)(x + 3)(x + 2)(x − 2) (3)
= x + 3x − 2 (1)
3.3
Write down the value(s) of x for which x + 3x − 2 is undefined, and explain why.
(3)
x = 2 (2) — the denominator becomes 0, and division by 0 is undefined (1)
3.4
Explain why you may NOT cancel the x² terms in x² + 5x + 6x² − 4 before factorising.
(3)
Only FACTORS may be cancelled (1)
Before factorising, x² is a TERM in a sum, not a factor (1)
Cancelling terms changes the value of the expression (1)

TOTAL: 35 marks

This question paper consists of 3 questions.