Edexcel P1 2018 Specimen — Question 3 6 marks

Exam BoardEdexcel
ModuleP1 (Pure Mathematics 1)
Year2018
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimultaneous equations
TypeLine intersecting general conic
DifficultyModerate -0.3 This is a standard simultaneous equations question combining a linear and quadratic equation. It requires substitution and solving a quadratic, which are routine A-level techniques. The algebra is straightforward with no unusual complications, making it slightly easier than average but still requiring multiple steps for the 6 marks available.
Spec1.02c Simultaneous equations: two variables by elimination and substitution
Solve the simultaneous equations $$y + 4x + 1 = 0$$ $$y^2 + 5x^2 + 2y = 0$$ [6]
Question 3:
AnswerMarks
3y (cid:32) − (cid:23)x − 1
(cid:159) (−(cid:23)x − 1(cid:12)2 + 5x2 + 2x = 0Attempts to makes y the subject of the
linear equation and substitutes into
AnswerMarks Guidance
the other equation.M1
21x2 + 10x + 1 = 0Correct 3 term quadratic A1
(7x+1)(3x+1) = 0 (cid:159)(cid:11)x(cid:32)(cid:12)(cid:16)1 , (cid:16)1
AnswerMarks
7 3dM1: Solves a 3 term quadratic by
the usual rulesdM1A1
1 1
A1: (x = )(cid:3)(cid:16) ,(cid:16) (cid:3)
7 3
3 1
y = – ,
AnswerMarks
7 3M1: Substitutes to find at least one y
valueM1 A1
3 1
A1: y = (cid:16) ,
7 3
(6)
Alternative
1 1
x(cid:32)(cid:16) y(cid:16)
4 4
(cid:167) 1 1(cid:183) 2 (cid:167) 1 1(cid:183)
(cid:159) y2 (cid:14)5 (cid:168)(cid:16) y(cid:16) (cid:184) (cid:14)2 (cid:168)(cid:16) y(cid:16) (cid:184)(cid:32)0
AnswerMarks
(cid:169) 4 4(cid:185) (cid:169) 4 4(cid:185)Attempts to makes x the subject of the
linear equation and substitutes into
AnswerMarks
the other equation.M1
21 1 3
y2 (cid:14) y(cid:16) (cid:32)0
16 8 16
AnswerMarks Guidance
(21y2 (cid:14)2y(cid:16)3(cid:32)0)Correct 3 term quadratic A1
3 1
(7y(cid:14)3)(3y(cid:16)1)(cid:32)0(cid:159)(y(cid:32))(cid:16) ,
AnswerMarks Guidance
7 3Solves a 3 term quadratic dM1
3 1
(cid:11)(cid:3)y (cid:32)(cid:3)(cid:12)(cid:16) , (cid:3)
AnswerMarks
7 3A1
1 1
x(cid:32)(cid:16) ,(cid:16)
AnswerMarks
7 3Substitutes to find at least one x
value.M1
1 1
x=(cid:16) ,(cid:16) (cid:3)(cid:3)
AnswerMarks
7 3A1
(6)
(6 marks)
AnswerMarks Guidance
QuestionScheme Marks 
Question 3:
3 | y (cid:32) − (cid:23)x − 1
(cid:159) (−(cid:23)x − 1(cid:12)2 + 5x2 + 2x = 0 | Attempts to makes y the subject of the
linear equation and substitutes into
the other equation. | M1
21x2 + 10x + 1 = 0 | Correct 3 term quadratic | A1
(7x+1)(3x+1) = 0 (cid:159)(cid:11)x(cid:32)(cid:12)(cid:16)1 , (cid:16)1
7 3 | dM1: Solves a 3 term quadratic by
the usual rules | dM1A1
1 1
A1: (x = )(cid:3)(cid:16) ,(cid:16) (cid:3)
7 3
3 1
y = – ,
7 3 | M1: Substitutes to find at least one y
value | M1 A1
3 1
A1: y = (cid:16) ,
7 3
(6)
Alternative
1 1
x(cid:32)(cid:16) y(cid:16)
4 4
(cid:167) 1 1(cid:183) 2 (cid:167) 1 1(cid:183)
(cid:159) y2 (cid:14)5 (cid:168)(cid:16) y(cid:16) (cid:184) (cid:14)2 (cid:168)(cid:16) y(cid:16) (cid:184)(cid:32)0
(cid:169) 4 4(cid:185) (cid:169) 4 4(cid:185) | Attempts to makes x the subject of the
linear equation and substitutes into
the other equation. | M1
21 1 3
y2 (cid:14) y(cid:16) (cid:32)0
16 8 16
(21y2 (cid:14)2y(cid:16)3(cid:32)0) | Correct 3 term quadratic | A1
3 1
(7y(cid:14)3)(3y(cid:16)1)(cid:32)0(cid:159)(y(cid:32))(cid:16) ,
7 3 | Solves a 3 term quadratic | dM1
3 1
(cid:11)(cid:3)y (cid:32)(cid:3)(cid:12)(cid:16) , (cid:3)
7 3 | A1
1 1
x(cid:32)(cid:16) ,(cid:16)
7 3 | Substitutes to find at least one x
value. | M1
1 1
x=(cid:16) ,(cid:16) (cid:3)(cid:3)
7 3 | A1
(6)
(6 marks)
Question | Scheme | Marks
Solve the simultaneous equations
$$y + 4x + 1 = 0$$
$$y^2 + 5x^2 + 2y = 0$$
[6]

\hfill \mbox{\textit{Edexcel P1 2018 Q3 [6]}}
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