0580/42 Feb/Mar 2024, Question 5: Simultaneous Equations by Elimination

Part (ai) · 2 marks

Factorise the quadratic expression x² − x − 12 into two brackets.

1. 1Two numbers with product −12 and sum −1: −4 and 3
2. 2x² − x − 12 = (x − 4)(x + 3)

Answer(x−4)(x+3)

Part (aii) · 2 marks

Simplify the fraction with numerator x² − 16 and denominator x² − x − 12.

1. 1x² − 16 = (x − 4)(x + 4)
2. 2(x − 4)(x + 4)/((x − 4)(x + 3))
3. 3= (x + 4)/(x + 3)

Answer(x+ 4)/(x+3)

Part (b) · 3 marks

Simplify the difference of squares (2x − 3)² − (x + 1)².

1. 1(2x − 3)² = 4x² − 6x − 6x + 9 = 4x² − 12x + 9
2. 2(x + 1)² = x² + 2x + 1
3. 34x² − 12x + 9 − x² − 2x − 1
4. 4= 3x² − 14x + 8

Answer3x²−14x + 8 or (x −4)(3x −2)

Part (c) · 4 marks

Combine (2x + 4)/(x + 1) − x/(x − 3) into one simplified fraction.

1. 1Common denominator (x + 1)(x − 3)
2. 2[(2x + 4)(x − 3) − x(x + 1)]/((x + 1)(x − 3))
3. 3= (2x² − 6x + 4x − 12 − x² − x)/((x + 1)(x − 3))
4. 4= (x² − 3x − 12)/((x + 1)(x − 3))

Answer(x²−3x−12)/((x+1)(x−3)) or (x²−3x −12)/(x²−2x −3)

Part (d) · 3 marks

Expand and simplify the product (x − 3)(x − 5)(2x + 1).

1. 1(x − 3)(x − 5) = x² − 8x + 15
2. 2(x² − 8x + 15)(2x + 1) = 2x³ + x² − 16x² − 8x + 30x + 15
3. 3= 2x³ − 15x² + 22x + 15

Answer2x³ − 15x² + 22x + 15

Part (e) · 6 marks

Solve x − 3y = 13 and 2x² − 9y = 116 simultaneously, showing all working.

1. 1From the first equation: 3y = x − 13, so 9y = 3(x − 13)
2. 22x² − 3(x − 13) = 116
3. 32x² − 3x − 77 = 0
4. 4(x − 7)(2x + 11) = 0, so x = 7 or x = −5 1/2
5. 5y = (x − 13)/3: x = 7 gives y = −2; x = −5 1/2 gives y = −18.5/3 = −6 1/6
6. 6x = 7, y = −2 and x = −5 1/2, y = −6 1/6

Answerx = 7, y = −2 and x = −5 1/2, y = −6 1/6

More past-paper questions on simultaneous equations by elimination

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