∑ DANEMATHICS
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Grade 11 · Equations & Inequalities
Simultaneous Equations: One Linear, One Quadratic (Grade 11)
MARKING GUIDELINE
Marks
32
Duration
50 minutes
Questions
3
Name:
Class:
Date:
Mark
/ 32
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.
- 1.1Solve simultaneously: y = x + 1 and y = x² − 1(1)A)(2 ; 3) and (−1 ; 0)✓B)(2 ; 3) onlyC)(−1 ; 0) onlyD)(2 ; 3) and (1 ; 2)Answer: A — x + 1 = x² − 1 gives x² − x − 2 = (x − 2)(x + 1) = 0.
- B — a line can cut a parabola TWICE — the second point was dropped
- C — only one of the two solutions was given
- D — x = 1 does not satisfy x² − x − 2 = 0
- 1.2For one LINEAR and one QUADRATIC equation, the standard method is to …(1)A)make a variable the subject of the LINEAR equation, then substitute✓B)make a variable the subject of the QUADRATIC equation, then substituteC)add the two equationsD)subtract the two equationsAnswer: A — The linear equation rearranges cleanly; the quadratic one does not.
- B — rearranging the quadratic brings in a square root and both of its signs
- C — adding works only for two LINEAR equations
- D — subtracting works only for two LINEAR equations
- 1.3Solve simultaneously: y = 2x and x² + y² = 20(1)A)(2 ; 4) and (−2 ; −4)✓B)(2 ; 4) onlyC)(2 ; 4) and (−2 ; 4)D)(4 ; 2) and (−4 ; −2)Answer: A — Substituting gives 5x² = 20, so x = ±2 and y = 2x follows.
- B — x² = 4 has TWO solutions, so there are two points
- C — y = 2x, so a negative x gives a NEGATIVE y
- D — the coordinates are the wrong way round
- 1.4How many solutions can a straight line and a parabola have?(1)A)0, 1 or 2✓B)always 2C)always 1D)0 or 1 onlyAnswer: A — The line may miss the parabola, touch it as a tangent, or cut it twice.
- B — a line that misses the parabola gives no solution at all
- C — a line usually cuts a parabola twice
- D — a line can cut a parabola twice
- 1.5Solve simultaneously: x + y = 5 and xy = 6(1)A)(2 ; 3) and (3 ; 2)✓B)(2 ; 3) onlyC)(1 ; 6) and (6 ; 1)D)(−2 ; −3) and (−3 ; −2)Answer: A — y = 5 − x gives x² − 5x + 6 = 0, so x = 2 or x = 3.
- B — both roots give a valid point, so there are TWO solutions
- C — 1 × 6 = 6, but 1 + 6 = 7, not 5
- D — (−2)(−3) = 6, but −2 + −3 = −5, not +5
- 1.6If substituting produces a quadratic with a NEGATIVE discriminant, then …(1)A)the line and the curve do not meet✓B)the line is a tangentC)there are two points of intersectionD)there are infinitely many solutionsAnswer: A — No real roots means no real point of intersection.
- B — a tangent gives Δ = 0
- C — two points need Δ > 0
- D — a line and a curve never share infinitely many points
- 1.7If substituting produces a quadratic with Δ = 0, the line is …(1)A)a tangent to the curve✓B)a secant cutting it twiceC)parallel to the curveD)missing the curve entirelyAnswer: A — One repeated root means the line touches the curve at exactly one point.
- B — cutting twice needs Δ > 0
- C — 'parallel' has no meaning for a line and a parabola
- D — missing it entirely needs Δ < 0
- 1.8Solve simultaneously: y = x − 1 and x² + y² = 25(1)A)(4 ; 3) and (−3 ; −4)✓B)(4 ; 3) onlyC)(3 ; 4) and (−4 ; −3)D)(4 ; 3) and (3 ; 4)Answer: A — Substituting gives x² − x − 12 = 0, so x = 4 or x = −3, and y = x − 1.
- B — the quadratic has TWO roots, so there are two points
- C — the coordinates are the wrong way round
- D — (3 ; 4) does not satisfy y = x − 1
- 1.9Two numbers have a product of 24 and a sum of 11. The numbers are …(1)A)3 and 8✓B)4 and 6C)2 and 12D)1 and 24Answer: A — x² − 11x + 24 = 0 gives x = 3 or x = 8, and 3 + 8 = 11.
- B — 4 × 6 = 24, but 4 + 6 = 10, not 11
- C — 2 × 12 = 24, but 2 + 12 = 14, not 11
- D — 1 × 24 = 24, but 1 + 24 = 25, not 11
- 1.10A rectangle has perimeter 26 cm and area 40 cm². Its dimensions are …(1)A)5 cm by 8 cm✓B)4 cm by 10 cmC)2 cm by 20 cmD)6 cm by 7 cmAnswer: A — l + b = 13 and lb = 40 give x² − 13x + 40 = 0, so the sides are 5 and 8.
- B — 4 × 10 = 40, but that perimeter is 28, not 26
- C — 2 × 20 = 40, but that perimeter is 44, not 26
- D — 6 + 7 = 13 gives the right perimeter, but 6 × 7 = 42, not 40
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
[10 MARKS]Solve the following simultaneously. Show ALL your working.
- 2.1y = x + 1 and y = x2 − 3x + 4(5)x + 1 = x2 − 3x + 4 (1)
x2 − 4x + 3 = 0 (1)
(x − 1)(x − 3) = 0 (1)
x = 1 gives y = 2 (1); x = 3 gives y = 4 (1) - 2.2x + y = 5 and x2 + y2 = 17(5)y = 5 − x (1)
x2 + (5 − x)2 = 17 (1)
2x2 − 10x + 8 = 0 (1)
(x − 1)(x − 4) = 0 (1)
(1 ; 4) and (4 ; 1) (1)
Question 3
[12 MARKS]Answer the questions below.
- 3.1Solve simultaneously: y − x = 2 and xy = 8(5)y = x + 2 (1)
x(x + 2) = 8 (1)
x2 + 2x − 8 = 0 (1)
(x + 4)(x − 2) = 0 (1)
(2 ; 4) and (−4 ; −2) (1) - 3.2Explain which of the two equations should always be rearranged first, and why.(2)The LINEAR one (1) — making a variable its subject introduces no squares, so the substitution stays simple (1)
- 3.3The line y = x + c cuts the parabola y = x2 at two points. Determine the values of c for which this happens.(5)x2 = x + c gives x2 − x − c = 0 (2)
Two points means Δ > 0 (1)
1 + 4c > 0 (1)
c > −14 (1)
TOTAL: 32 marks
This question paper consists of 3 questions.