Download Homework on analysing questionnaires – grade C

The Robert Smyth School
Mathematics Faculty
Topic 9
Solving Equations 2
Innovation & excellence
HW4 – Grade A/A*
Simultaneous Equations – 1 Linear 1 Quadratic
1.
Solve the simultaneous equations
y=x+2
y = 3x2
You must show your working.
Do not use trial and improvement.
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Answer .......................................................................................................
(Total 5 marks)
2.
Solve the simultaneous equations
y = 3x2
5x + y = 2
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Answer x =....................................................... y =................................................................
(Total 5 marks)
The Robert Smyth School
1
The Robert Smyth School
Mathematics Faculty
Topic 9
Solving Equations 2
Innovation & excellence
3.
Solve the simultaneous equations.
y=x+7
x2 + y2 = 25
YOU must show your working.
Do not use trial and improvement.
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Answer ..................................................................
(Total 7 marks)
4.
A straight line has the equation
y = 2x – 3
A curve has the equation
y2 = 8x – 16
(a)
Solve these simultaneous equations to find any points of intersection of the line and
the curve.
Do not use trial and improvement.
You must show all your working.
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The Robert Smyth School
2
The Robert Smyth School
Mathematics Faculty
Topic 9
Solving Equations 2
Innovation & excellence
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Answer …………………………………
(5)
(b)
Here are three sketches showing the curve y2 = 8x – 16 and three possible positions of
the line y = 2x – 3
y
x
Sketch 1
–3
y
Sketch 2
x
–3
y
Sketch 3
–3
x
Which is the correct sketch?
You must explain your answer.
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(2)
(Total 7 marks)
The Robert Smyth School
3
The Robert Smyth School
Mathematics Faculty
Topic 9
Solving Equations 2
Innovation & excellence
1.
3x2 = x + 2
M1
3x2 – x – 2 = 0
A1
(3x + 2)(x – 1) = 0
x = 1 or –
y = 3 or
2
3
M1
or [1 ± (1 – –24)]/6
A1
Accept –0.66 or –0.67
4
3
A1 ft
Must match appropriate values of y with x
x = 1, y = 3 without working ...SC1
[5]
2.
2 – 5x = 3x2
M1
2–
or y = 3 
 5
y


2
3x2 + 5x – 2 = 0
M1
3y2 – 37y + 12 = 0
(3x – 1)(x + 2)
M1
(3y – 1)(y – 12)
x=
1
or – 2
3
A1
y=
y=
1
or 12
3
1
or 12
3
A1
x=
1
or – 2
3
[5]
3.
x2 + (x + 7)2 = 25
or (y – 7)2 + y2 = 25
For substitution
M1
x2 + 14x + 49
or y2 – 14y + 49
M1
2
2
For expansion of (y – 7) or (x + 7)
(at least 3 correct terms)
2x2 + 14x + 24 = 0
or 2y2 – 14y + 24 = 0
Complete simplification and all on one side of equation
The Robert Smyth School
M1 dep
4
The Robert Smyth School
Mathematics Faculty
Topic 9
Solving Equations 2
Innovation & excellence
Dependent on both previous marks
(x + 4)(x + 3) = 0
or (y – 4)(y – 3) = 0
Or (2x + 8)(x + 3) = 0
Or (x + 4)(2x + 6) = 0
Or (2y – 8)(y – 3) = 0
Or (y – 4)(2y – 6) = 0
A1
7 1
2
– 7 1
Or x =
2
oe
Or y =
x = – 4and x = – 3
or y = (+)4 and y = (+)3
Or 1 correct pair
A1
y = (+)3 and y = (+)4
or x = – 4 and x = – 3
A1
Both correct pairings
x = –4, y = (+)3 SC1
x = –3, y = (+)4 SC1
Note: Do not award SC marks from clearly incorrect working
A1
[7]
4.
(a)
(b)
(2x – 3)2 = 4x2 – 6x - 6x + 9
condone one error
M1
2
4x + 9 is two errors
4x2 – 12x + 9 = 8x – 16
or 4x2 – 20x +25 (= 0)
for equating expressions and/or simplifying
this must lead to a quadratic equation
Ml
(2x – 5)(2x – 5) (– 0)
ft from their quadratic equation (if 'formula' used, substitution
must be completely correct)
M1
x = 2.5
A1
y=2
A1
Only one solution so straight line
must be a tangent to the curve
Hence sketch 2
ft from their solution(s) to (a)
clear solution(s) in (a)  ‘correct’ sketch in (b) can earn
B1 (no explanation) or B2 (with explanation)
B2ft
[7]
alternatively
The Robert Smyth School
5
The Robert Smyth School
Mathematics Faculty
Topic 9
Solving Equations 2
Innovation & excellence
(a)
x
 y  3 , y  3  2x, 4 y  8x 12
2
M1
oe for setting up attempt to eliminate x
(b)
y2 – 4y + 4 (= 0)
oe condone one error
this must lead to a quadratic equation
M1
(y – 2)(y – 2)(–0)
ft from their quadratic equation (if ‘formula’ used,
substitution must be completely correct)
M1
y=2
A1
x = 2.5
A1
Only one solution so straight line
must be a tangent to the curve
Hence sketch 2
ft from their solution(s) to (a)
clear solution(s) in (a)  ‘correct’ sketch in (b) can earn
B1 (no explanation) or B2 (with explanation)
B2ft
[7]
The Robert Smyth School
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