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```P1 Chapter 3 :: Equations and
Inequalities
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Chapter Overview
There is little new content in this chapter since GCSE.
1:: Simultaneous Equations
2:: Simultaneous Equations using Graphs
Solve:
Find the points of intersection of
π¦ = 3π₯ 2 β 2π₯ + 4 and 7π₯ + π¦ + 3 = 0
π₯ + π¦ = 11
π₯π¦ = 30
NEW! (since GCSE)
You may have to use the discriminant to show that
the two graphs have no points of intersection.
3:: Solving Inequalities
4:: Sketching Inequalities
Find the set of values of π₯ for
which:
π₯ 2 β 11π₯ + 24 < 0
Sketch the region that
satisfies the inequalities:
2π¦ + π₯ < 14
π¦ β₯ π₯ 2 β 3π₯ β 4
NEW! (since GCSE, and new to A Level 2017+)
Use set notation to represent solutions to
inequalities.
Solutions sets
The solution(s) to an equation may be:
A single value:
2π₯ + 1 = 5
Multiple values:
π₯ 2 + 3π₯ + 2 = 0
An infinitely large
set of values:
π₯>3
No (real) values!
π₯ 2 = β1
Every value!
π₯2 + π₯ = π₯ π₯ β 1
The point is that you shouldnβt think of the solution to an equation/inequality as
an βanswerβ, but a set of values, which might just be a set of 1 value (known as a
singleton set), a set of no values (i.e. the empty set β), or an infinite set (in the
last example above, this was β)
! The solutions to an equation are known as the solution set.
Solutions sets
For simultaneous equations, the same is true, except each βsolutionβ in the
solution set is an assignment to multiple variables.
All equations have to be satisfied at the same time, i.e. βsimultaneouslyβ.
Scenario
A single solution:
Two solutions:
Example
π₯+π¦ =9
π₯βπ¦ =1
π₯2
π¦2
+
= 10
π₯+π¦ =4
No solutions:
π₯+π¦ =1
π₯+π¦ =3
Infinitely large set
of solutions:
π₯+π¦ =1
2π₯ + 2π¦ = 2
Solution Set
Solution 1: π = π, π = π
To be precise here, the solution set is of
size 1, but this solution is an assignment
to multiple variables, i.e. a pair of values.
Solution 1: π = π, π = π
Solution 2: π = π, π = π
This time we have two solutions,
each an π₯, π¦ pair.
The solution set is empty, i.e.
β, as both equation canβt be
satisfied at the same time.
Solution 1: π = π, π = π
Solution 2: π = π, π = π
Solution 3: π = π, π = βπ
Solution 4: π = π. π, π = π. π
β¦ Infinite possibilities!
Textbook Error Pg39:
βLinear simultaneous
equations in two
unknowns have one set
of values that will make
a pair of equations true
at the same time.β
There are two separate
errors in this statement
β Iβll let you work out
what!
1 :: Simultaneous Equations
Recap!
Solve the simultaneous equations:
Solve the simultaneous equations:
3π₯ + π¦ = 8
2π₯ β 3π¦ = 9
π₯ + 2π¦ = 3
π₯ 2 + 3π₯π¦ = 10
We can either use substitution (i.e.
making π₯ or π¦ the subject of one
equation, and substituting it into the
other) or elimination, but the latter is
easier for linear equations.
ππ + ππ = ππ
ππ β ππ = π
Adding the two equations to
βeliminateβ π:
πππ = ππ β π = π
Substituting into first equation:
ππ + ππ = ππ β π = βπ
We canβt use elimination this time as nothing
would cancel.
(1) Rearrange linear equation to make π₯ or π¦
the subject.
(2) Substitute into quadratic equation and solve.
π = π β ππ
Substitute into other equation:
π β ππ π + ππ π β ππ = ππ
β¦ πππ + ππ + π = π
ππ + π π + π = π
π
π=β
β π=π
π
π = βπ β π = π
Solve the simultaneous equations:
3π₯ 2 + π¦ 2 = 21
π¦ =π₯+1
3π₯ 2 + π₯ + 1 2 = 21
3π₯ 2 + π₯ 2 + 2π₯ + 1 = 21
4π₯ 2 + 2π₯ β 20 = 0
2π₯ 2 + π₯ β 10 = 0
2π₯ + 5 π₯ β 2 = 0
5
π₯ = β ππ π₯ = 2
2
3
π¦ = β ππ π¦ = 3
2
Exercise 3A/B
Pearson Pure Mathematics Year 1/AS
Pages 40, 41
Extension
2 [STEP 2010 Q1] Given that
5π₯ 2 + 2π¦ 2 β 6π₯π¦ + 4π₯ β 4π¦
1 [MAT 2012 1G] There are
β‘ π π₯ β π¦ + 2 2 + π ππ₯ + π¦ 2 + π
positive real numbers π₯ and
a) Find the values of π, π, π, π.
π¦ which solve the equations
b) Solve the simultaneous equations:
2π₯ + ππ¦ = 4,
5π₯ 2 + 2π¦ 2 β 6π₯π¦ + 4π₯ β 4π¦ = 9,
π₯+π¦ =π
6π₯ 2 + 3π¦ 2 β 8π₯π¦ + 8π₯ β 8π¦ = 14
for:
(Hint: Can we use the same method in (a) to rewrite the second equation?)
A) All values of π;
a) Expanding RHS:
B) No values of π;
π + πππ ππ + π + π ππ + βππ + πππ ππ + πππ β πππ
C) π = 2 only;
+ (ππ + π)
D) Only π > β2
Comparing coefficients: π = π, π = π, π = βπ, π = βπ
If π = π then ππ + ππ = π and
π + π = π which are equivalent.
This would give an infinite
solution set, thus the answer is
C.
b) π β π + π π + βππ + π π β π = π
Using method in (a): π π β π + π π + βππ + π π β π = ππ
Subtracting yields π β ππ = ±π and π β π + π = ±π
We have to consider each of 4 possibilities.
Final solution set: π = βπ, π = βπ ππ π = π, π = π
or π = π, π = π ππ π = π, π = ππ
Simultaneous Equations and Graphs
Recall that a line with a given
equation is the set of all points
that satisfy the equation.
i.e. It is a graphical representation
of the solution set where each
(π₯, π¦) point represents each of the
solutions π₯ and π¦ to the equation.
e.g. π₯ = 3, π¦ = 7
Now suppose we introduced a second simultaneous equation:
π¦ = 2π₯ + 1
π₯+π¦ =5
The second line again consists of all points (π₯, π¦) which satisfy the
equation. So what point must satisfy both equations simultaneously?
The point of intersection!
Example
a) On the same axes, draw the graphs of:
2π₯ + π¦ = 3
π¦ = π₯ 2 β 3π₯ + 1
b) Use your graph to write down the
solutions to the simultaneous equations.
π = βπ, π = π ππ
π = π, π = βπ
(We could always substitute into the original
equations to check they work)
c) What algebraic method (perhaps
thinking about the previous chapter), could
we have used to show the graphs would
have intersected twice?
Substituting linear equation into quadratic:
π = π β ππ
β΄ π β ππ = ππ β ππ + π
ππ β π β π = π
Since there were two points of intersection, the
equation must have two distinct solutions. Thus
ππ β πππ > π
π = π, π = βπ, π = βπ
π+π=π>π
Thus the quadratic has two distinct solutions,
i.e. we have two points of intersection.
Another Example
a) On the same axes, draw the graphs of:
π¦ = 2π₯ β 2
π¦ = π₯ 2 + 4π₯ + 1
b) Prove algebraically that the lines
never meet.
When we try to solve
simultaneously by substitution,
the equation must have no
solutions.
ππ + ππ + π = ππ β π
ππ + ππ + π = π
π = π, π = π, π = π
ππ β πππ = π β ππ = βπ
βπ < π therefore no solutions,
and therefore no points of
intersection.
Final Example
The line with equation π¦ = 2π₯ + 1 meets the curve with equation
ππ₯ 2 + 2π¦ + π β 2 = 0 at exactly one point. Given that π is a positive constant:
a) Find the value of π.
b) For this value of π, find the coordinates of this point of intersection.
a
Substituting:
ππ₯ 2 + 2 2π₯ + 1 + π β 2 = 0
ππ₯ 2 + 4π₯ + 2 + π β 2 = 0
ππ₯ 2 + 4π₯ + π = 0
Since one point of intersection, equation has one solution, so π2 β 4ππ = 0.
π = π, π = 4, π = π
16 β 4π 2 = 0
π = ±2
But π is positive so π = 2.
b
When π = 2,
2π₯ 2
We can breathe a sigh of
relief as we were expecting
one solution only.
+ 4π₯ + 2 = 0
π₯ 2 + 2π₯ + 1 = 0
π₯ + 1 2 = 0 β π₯ = β1
π¦ = 2 β1 + 1 = β1 β (β1, β1)
Exercise 3C
Pearson Pure Mathematics Year 1/AS
Page 45
Set Builder Notation
Recall that a set is a collection of values such that:
a) The order of values does not matter.
b) There are no duplicates.
Recap from GCSE:
Froflection: Sets seem
sensible for listing solutions
to an equation, as the order
doesnβt matter.
β’ We use curly braces to list the values in a set, e.g. π΄ = 1,4,6,7
β’ If π΄ and π΅ are sets then π΄ β© π΅ is the intersection of π΄ and π΅, giving a set which
has the elements in π΄ and π΅.
β’ π΄ βͺ π΅ is the union of π΄ and π΅, giving a set which has the elements in π΄ or in π΅.
β’ β is the empty set, i.e. the set with nothing in it.
β’ Sets can also be infinitely large. β is the set of natural numbers (all positive
integers), β€ is the set of all integers (including negative numbers and 0) and β
is the set of all real numbers (including all possible decimals).
β’ We write π₯ β π΄ to mean βπ₯ is a member of the set Aβ. So π₯ β β would mean
βπ₯ is a real numberβ.
1,2,3 β© 3,4,5 = π
1,2,3 βͺ 3,4,5 = π, π, π, π, π
=β
Set Builder Notation
It is possible to construct sets without having to
explicitly list its values. We use:
The | or : means βsuch thatβ.
or
ππ₯ππ ππππππ‘πππ }
{ππ₯ππ βΆ ππππππ‘πππ }
Can you guess what sets the following give?
(In words βAll numbers 2π₯ such that π₯ is an integer)
2π₯ βΆ π₯ β β€ = {0,2, β2,4, β4,6, β6, β¦ }
i.e. The set of all
even numbers!
2π₯ βΆ π₯ β β = {2,4,8,16,32, β¦ }
π₯π¦: π₯, π¦ πππ πππππ = {4,6,10,14,15, β¦ }
i.e. All possible products
of two primes.
We previously talked about βsolutions setsβ, so set builder notation is
very useful for specifying the set of solutions!
Set Builder Notation
Can you use set builder notation to specify the following sets?
All odd numbers.
{2π₯ + 1 βΆ π₯ β β€}
All (real) numbers
greater than 5.
{π₯: π₯ > 5}
All (real) numbers less
than 5 or greater than 7.
All (real) numbers
between 5 and 7 inclusive.
Technically it should be {π₯: π₯ > 5, π₯ β β} but
the π₯ > 5 by default implies real numbers
greater than 5.
π₯: π₯ < 5 βͺ {π₯: π₯ > 7}
We combine the two sets together.
π₯: 5 β€ π₯ β€ 7
While we could technically write
π₯: π₯ β₯ 5 β© {π₯: π₯ β€ 7}, we tend to write multiple
required conditions within the same set.
Recap of linear inequalities
Inequality
Solution Set
2π₯ + 1 > 5
{π₯ βΆ π₯ > 2}
3 π₯ β 5 β₯ 5 β 2(π₯ β 8)
{π₯ βΆ π₯ β₯ 7.2}
{π₯ βΆ π₯ β€ β2}
βπ₯ β₯ 2
Fro Note: Multiplying or both sides
of an inequality by a negative
number reverses the direction.
Combining Inequalities:
If π₯ < 3 and 2 β€ π₯ < 4, what is the combined solution set?
2
3
4
2β€π₯<3
If both inequalities have to be satisfied,
we have to be on both lines. Place your
finger vertically and scan across.
RECAP :: Solving Quadratic Inequalities
Solve π₯ 2 + 2π₯ β 15 > 0
Step 1: Get 0 on one side
π₯+5 π₯β3 >0
Step 2: Factorise
Step 3: Sketch and reason
π¦
π¦ = (π₯ + 5)(π₯ β 3)
β5
3
π₯
Since we sketched π¦ = (π₯ + 5)(π₯ β 3)
weβre interested where π¦ > 0, i.e. the
parts of the line where the π¦ value is
positive.
Click to Fro-Bolden >
What can you say about
the π₯ values of points in
this region?
π < βπ
What can you say about
the π₯ values of points in
this region?
π>π
π₯: π₯ < β5 βͺ {π₯: π₯ > 3}
Fro Note: If the π¦ value is βstrictlyβ greater than 0,
i.e. > 0, then the π₯ value is strictly less than -5. So
the < vs β€ must match the original question.
Solve π₯ 2 + 2π₯ β 15 β€ 0
Step 1: Get 0 on one side
π₯+5 π₯β3 β€0
Step 2: Factorise
Step 3: Sketch and reason
π¦
π¦ = (π₯ + 5)(π₯ β 3)
β5
3
Again, what can we say about the π₯
value of any point in this region?
π₯
{π₯ βΆ β5 β€ π₯ β€ 3}
Bro Note: As discussed
previously, we need β€
rather than < to be
consistent with the
original inequality.
Further Examples
Solve ππ + ππ β₯ βπ
π₯ 2 + 5π₯ + 4 β₯ 0
π₯+4 π₯+1 β₯0
π¦ = (π₯ + 4)(π₯ + 1)
β4
Solve ππ < π
π₯2 β 9 < 0
π₯+3 π₯β3 <0
π¦
β1
π¦
π₯
π₯ β€ β4 or π₯ β₯ β1
Fro Note: The most common error Iβve seen students
make with quadratic inequalities is to skip the βsketch
stepβ. Sodβs Law states that even though you have a 50%
chance of getting it right without a sketch (presuming
youβve factorised correctly), you will get it wrong.
β3
π¦ = (π₯ + 3)(π₯ β 3)
3
π₯
β3 < π₯ < 3
βUse of Technologyβ Monkey says:
When Iβm not busy flinging poo at other monkeys, I use
the quadratic inequality solver on my ClassWiz. Just go to
Menu β Inequalities, then choose βorder 2β (i.e. quadratic)
Edexcel C1 May 2010 Q3
Edexcel C1 June 2008 Q8
Fro Note: What often confuses
students is that the original
equation has no solutions, but the
inequality π 2 + 8π < 0 did have
solutions. But think carefully what
weβve done: Weβve found the
solutions for π that result in the
original equation not having any
solutions for π₯. These are different
variables, so have different
solutions sets, even if the solution
set of π influences the solution set
of π₯.
Deal with inequalities with a division by π₯
Spec Note: This is an
example in the textbook,
although it is ambiguous
whether this type of
question is in the new
specification. Dealing with
an π₯ in the denominator
within an inequality is a
skill previously in the old
Further Pure 2 module.
But you never really know!
6
π₯
Find the set of values for which > 2, π₯ β  0
Why canβt we just multiply both sides by π?
We earlier saw that multiplying by a negative number would flip the
inequality, but multiplying by a positive number would not. Since we
donβt know π₯, we donβt know whether the inequality would flip or not!
6
Once solution is to sketch π¦ = and π¦ = 2, find the points of
π₯
intersection and reason about the graph (see next section,
βInequalities on Graphsβ), but an easier way is to multiply both sides
by ππ , because it is guaranteed to be positive:
π¦
ππ > πππ
πππ β ππ < π
ππ β ππ < π
π πβπ <π
π<π<π
π¦ = π₯(π₯ β 3)
0
3
π₯
Exercise 3D/3E
Pearson Pure Mathematics Year 1/AS
Page 47-48, 50-51
Inequalities on Graphs
(New to the 2017 spec)
π¦
π¦ = (π₯ + 5)(π₯ β 3)
β5
π₯
3
When we solved quadratic
inequalities, e.g. π₯ + 5 π₯ β 3 > 0
We plotted π¦ = (π₯ + 5)(π₯ β 3) and
observed the values of π₯ for which
π¦ > 0.
Can we use a similar method when we
donβt have 0 on one side?
Example: πΏ1 has equation π¦ = 12 + 4π₯. πΏ2 has equation π¦ = π₯ 2 .
The diagram shows a sketch of πΏ1 and πΏ2 on the same axes.
a) Find the coordinates of π1 and π2 , the points of intersection.
b) Hence write down the solution to the inequality 12 + 4π₯ > π₯ 2 .
π¦
a Solve simultaneously to find points of intersection:
π2
π₯ 2 = 12 + 4π₯
π₯ 2 β 4π₯ β 12 = 0
π₯ = 6, π₯ = β2 β π¦ = 36, π¦ = 4
π1 6,36 ,
π2 β2,4
π1
β2
6
π₯
b When 12 + 4π₯ > π₯ 2 the πΏ1 graph is above the πΏ2
graph. This happens when β2 < π₯ < 6.
Inequality Regions
On graph paper, shade the region that satisfies the inequalities:
2π¦ + π₯ < 14
π¦ β₯ π₯ 2 β 3π₯ β 4
You did this at GCSE, the only difference here being that the graphs involved might
not be straight lines.
Step 2:
An inequality involving π₯ and π¦
represents a 2D region in space.
Identify the correct side of each
line each inequality represents.
π¦
β1
4
Fro Tip: To quickly sketch
2π¦ + π₯ = 14, consider what
happens when π₯ is 0 and
when π¦ is 0.
π₯
Click to Frosketch >
Fro Tip: Make sure π¦ is on the side where it is positive.
If π¦ is on the smaller side, youβre below the line.
If π¦ is on the greater side, youβre above the line.
Exercise 3F/3G
Pearson Pure Mathematics Year 1/AS
Page 53, 55
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