Equations


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Simultaneous Equations

a⁢x+b⁢y=c p⁢x+q⁢y=r

Parametric Equations

x=r⁢cosθ y=r⁢sinθ

Parametric Equations

Parametric equations are a set of explicit equations that represent a mapping that is not a function, i.e. a many-to-many mapping or a one-to-many mapping. These can be useful in solving problems using the methods known for handing functions.

A curve, C can be defined using a set of parametric equations x=u(θ) and y=v(θ). Each value of θ defines a point on C with coordinates (u(θ),v(θ)).

For a given set of parametric equations a cartesian equation can be formed by using substitution to eliminate the parameter.

For parametric equations x=u(θ) and y=v(θ) with Cartesian equation y=f(x):


Questions:

A level Edexcel Year 2 Text book chapter 8 page 197

Solution Bank

Parametric equations
(Care of alevelmaths.co.uk)

Parametric equations
(Care of www.physicsandmathstutor.co.uk)

Parametric Curves
(Care of www.physicsandmathstutor.co.uk)

Exam parametric equations
(Care of www.examsolutions.net)


(1)

x y C O
Figure 1.1

Figure 1.1 shows a sketch of the curve C with parametric equations

x=4⁢cos(t+π6),
y=2⁢sint,
0<t≤2π

Show that the Cartesian equation of C can be written in the form:

(x+y)2+a⁢y2 =b

Where a and b are constants to be found.


[ Solution 1 ]

x+y =4⁢cos(t+π6) +2⁢sint x+y =4⁢ ( cost⁢cosπ6 -sint⁢sinπ6 ) +2⁢sint x+y =4⁢ ( 32⁢cost -12⁢sint ) +2⁢sint x+y=2⁢3⁢cost (x+y)2 =12⁢cos2t (x+y)2 =12⁢ (1-sin2t) (x+y)2 =12-3⁢y2 (x+y)2+3⁢y2 =12 a=3, b=12

Further Reading:

Parametric Equations
(Care of alevelmaths.co.uk)

Exam Questions - Parametric Equations
(Care of www.examsolutions.net)


Next Steps:

Parametric Differentiation
(Care of revisionmaths.com)

Parametric Integration


Simultaneous Equations

Warm-up

Rearranging Formulae
(Care of mmerevise.co.uk)

Graphical
Textbook Exercise


Worked Examples:

Example 1:

Solve these two simultaneously using elimination:

y=3⁢x+7 4⁢x+y=14

Aligning the terms of each equation first

4⁢x+y=3⁢x+7 4⁢x+y=3⁢x+14

subtracting

-4⁢x+y=3⁢x-7

Simplifying

7=7⁢x x=1

Substituting this into the first equation in the question,

y=3×1+7=10

Checking by substituting both into the second equation.

4×1+10=14

Example 2:

Solve these two simultaneously:

3⁢x+5⁢y=13 5⁢x-y=3

Multiplying the second equation by five,

25⁢x-5⁢y=15

Then adding the third to the first equation,

28⁢x=28

So x=1

Substituting this result into the second equation gives y=2

Example 3:

Solve by substitution:

5⁢x+3⁢y=41 2⁢x+3⁢y=20

Making 3⁢y the subject of the first equation:

3⁢y=41-5⁢x

The substituting this into the second:

2⁢x+41-5⁢x=20 -3⁢x=-21 x=7

Substituting x=7 into the first equation:

5×7+3⁢y=41 3⁢y=6 y=2

Example 4:

Solve these two simultaneously:

x2+y2=20 x+y=6

Rearrange the second equation to give y as the subject

y=6-x

and substituting in the first equation and solving

x2+(6-x)2 =20 x2+36-12⁢x+x2 =20 x2-6⁢x+8=0
x=2y=4

or

x=4y=2

Questions and Further Materials

Simultaneous Equations
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Simultaneous equations
(care of www.mathsgenie.co.uk)

Elimination and Substitution, both linear
Practice Questions
Textbook Exercise
Advanced inc quadratic
Practice Questions 1
Practice Questions 2 [ MS ].
Textbook Exercise
3 Unknowns
Textbook Exercise
(care of corbettmaths.com)
Solving simultaneous equations
(care of www.mrbartonmaths.com)

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