Transcript ALGEBRA - Math4cxc

ALGEBRA
ALGEBRA(Topics)
• Algebra Basics- Addition, Subtraction and
Substitution.
• Algebra Basics 2- Multiplication, Expansion of
Brackets and Division
• Indices
• Equations and Inequalities
• Simultaneous Equations
• Simultaneous Equations in solving equations
ALGEBRA(Topics)
• Functions and Rearranging.
• Quadratic equations
ALGEBRA(Key Terms)
•
Substitution-replacing occurrences of some symbol or variable by a given value.
•
Evaluate-to determine or calculate the numerical value of
•
•
•
•
•
Variable-a quantity or function that may assume any given value or set of values
Constant- A quantity assumed to have a fixed value in a specified mathematical context
Simplify-to reduce (an equation, fraction, etc) to a simpler form by
cancellation of common factors, regrouping of terms in the same variable.
Coefficient-a number or quantity placed (generally) before and
multiplying another quantity, as 3 in the expression 3x.
=> means ‘this gives’
nb: means ‘note well’
•
Multiplication operator can be represented by a ‘*’ or a ‘
•
So…… 2x3=6, 2*2=4, 5 4=20
•
.
.
’ or a ‘x ‘ sign.
ALGEBRA(Basics)
• The four basic operations for addition,
subtraction, multiplication and division are
carried out on variables much like on numbers.
• One practical way to look at it is if I have a
variable ‘a’ and I am adding ‘2a’ to it. Imagine
a=‘apples'. Ask yourself if I have one apple and
someone gives me two how many do I have.
Questions in the exam normally come in the
form:a+2a
This gives 3a.
ALGEBRA(Basics)
• The same is done with subtraction
• If I have 5 mangoes and my grandmother
takes 3 mangoes how many do I have.
• Let ‘m’ be equal to mangoes(step 1)
• Represent it in a mathematical statement(step
2)
5m-3m=
Then Evaluate the statement 5m-3m=2m.(step3)
ALGEBRA(Basics)
• Sometimes Variables have negative coefficients.
• For example, You were playing cricket and you lost
the ball by hitting a massive ‘6’ you now owe one
of your friends a ball. Hence you have -balls(or
negative 1 balls).
• The next day your mother buys a three pack of
balls but you still owe your friend, this situation can
be represented in a mathematical statement as
-b+3b
This means you actually own only 2 of those
balls, sine you owe you friend. Therefore –b+3b=2b.
ALGEBRA(Basics)
• This statement can be rewritten as
3b+-b which is equal to 3b-b which is
equal to 2b.
ALGEBRA(Basics)
• When writing mathematical statements or
expressions variables of different types
cannot be combined or simplified further.
• If you have three balls and two apples you
can represent it as
3b+2a
where b=‘balls’ and a=‘apples’
ALGEBRA(Basics)
• You may be given an expression like eg.1
-b+3a+2b-4a
And asked to simplify the expression.
Bring similar(or like) variables together, that is
‘a’s with ‘a’s and ‘b’s with ‘b’s. Take whatever
sign is in front of them and rewrite the
statement. Do one variable first then the
next.(step 1)
-b+2b+3a-4a
ALGEBRA(Basics)
-b+2b+3a-4a
Again negative one ‘b’ plus two b , is one b
as if subtracting. Three ‘a’ minus four ‘a’ give
negative one ‘a’. So we can simplify the
expression as
b+-a which is equal to b-a
ALGEBRA(Basics)
• eg.2
3p-4q+6p+2q
Collect like terms with the same sign and
rewrite
=3p+6p-4q+2q
simplify(remember-4p+2q =2q+-4q =2q-4q =-2q)
=9p-2q
ALGEBRA(Basics)
• eg.3
5c-2d+7d-3c
Collect terms and rewrite with the same
signs in front of them
=5c-3c-2d+7d
Simplify(remember -2d+7d =7d+-2d =7d-2d =5d)
=2c+5d
ALGEBRA(Basics)
• Sometimes we are given values for our
variables we can Substitute the to get
actual values for our expressions.
• eg.1 Given x=3, y=2 Evalutate x+y
• Replace x with the value given for x and y
with the value given for y and rewrite(Step
1)
• NB: it is best to use brackets.
ALGEBRA(Basics)
•
•
•
•
Evaluate (Step 2) x+y
= (3)+(2)
=5
We will show why brackets are important further down
in this Topic.
eg.2 Given a=2 and b=1 Evaluate 2a+2b
Replace(Substitute) ‘a’ with 2 and ‘b’ with 1 in the
expression.(Step 1)
Expand Brackets and Evaluate(Step 2)
=>2(2)+2(1)
=4+2
=6
ALGEBRA(Basics)
eg.3 Given c=-1 and d=2 Evaluate c+3d
=(-1)+3(2)
=> -1+6 (NB:6+-1 = 6-1)
=5
eg.4 Given e=2 and f=-3 evaluate e-2f
Rewrite expression as it is just substituting the variables
=>(2)-2(-3)
=>2-(-6)
=8
Expand Brackets, then evaluate
We get 8 because if I take away some thing negative like debt it is as
if I have added to what you have. Imagine you owe your father $4 that
is -$4 you have. If your mother convinces him to clear the debt she has
taken away the -$4.
-$4-(-$4)=0 Like adding
-$4+$4=0
ALGEBRA(Basics)
• Algebra Basics Ends Here go to Algebra
Exercise section and do Basic revision
exercise.
ALGEBRA(Basics 2)
• There is a general sequence when working
with Algebraic expressions involving which
operations to conduct first. B.O.D.M.A.S
• B.O.D.M.A.S – Brackets (operated on
before),Division(then),Multiplication(then),
Addition(then),Subtraction.
ALGEBRA(Basics 2)
eg.1 Simplify 3(x+2)=
Expand brackets first by multiplying
everything inside the bracket by the
coefficient outside the bracket
3(x)+3(2)=
3x+6
ALGEBRA(Basics 2)
eg.2 Simplify 3y-2(y+4)=
According to BODMAS let us expand brackets first.
That is take the coefficient in front the bracket and
multiply it by everything inside the bracket. The
coefficient is -2 so by expanding the bracket we get
3y+-2(y)+-2(4) {nb: rewrite sign the bracket}
=>3y+-2y+-8
=>3y-2y-8
=>y-8
ALGEBRA(Basics 2)
eg.3 Simplify 3x-2(x+5)
Remember B.O.D.M.A.S!
=>3x+-2(x)+-2(5) {nb: rewrite sign in front of the
bracket}
=>3x-2x-10
=>x-10
ALGEBRA(Basics 2)
eg.4 Simplify 5(a-b)+2b
=>5(a)-5(b)+2b {nb: rewrite sign the bracket}
=>5a-5b+2b
=>5a-3b
ALGEBRA(Basics 2)
eg.4 Simplify 3(a-b)-a
=>3(a)-3(b)-a
=>3a-3b-a
By rearranging and collecting terms =>3a-a-3b
=>2a-3b{Remember when collecting terms
keep the sign in front of them}
ALGEBRA(Basics 2)
Variables are multiplied similarly to numbers
a*b=ab. {Note * means multiply}
Eg.5Expand the brackets a(b-c)=
a(b)-a(c)=
ab-ac
Any variable by itself is equal to the variable
squared eg b*b=b2
Eg.6 2b(a-b)=
2ab-2b2
ALGEBRA(Basics 2)
When dividing we cancel similar terms and divide
the numbers (coefficients)
Eg.7
2𝑎 2𝑎
=
𝑎 𝑎
Eg.8
Eg.9
6𝑏
2𝑏
=
4𝑎𝑏
2𝑎
6𝑏
2𝑏
=
=2
6
2
= =3
4𝑏
=
2
2b
ALGEBRA(Basics 2)(Key Terms)
Numerator- the number at the top of a
fraction and so is being divided.
Denominator- the number at the bottom of
the fraction which is dividing the numerator.
LCM-Lowest Common Multiple.
ALGEBRA(Basics 2)
Remember with numbers when adding two fractions with
different denominators we found the LCM. For example
3
4
2
5
3
4
2
+
The LCM is found by multiplying the denominators (4*5=20) this is
the denominator of the resulting fraction.
+
5
We want to convert both original fractions to equal values with
the denominator of the LCM.
20
Step 1. divide the LCM by the denominator of the first fraction
the multiply the quotient by the numerator of that fraction.
Step 2.Repeat for the second fraction.
ALGEBRA(Basics 2)
This gives 15 and 8.
15+8 23
=
20
20
Same process would be done with a ‘-’ sign.
This section was simply revision.
Algebraic fractions undergo the same
operations…..
ALGEBRA(Basics 2)
Ex.10 Simply
2𝑎
𝑏
+
3𝑐
𝑑
Step 1. Find LCM by multiplying
denominators, b*d=bd
Step 2. Divide denominators of both fractions
and multiply the quotient by their
numerators.
Step 3. Place the result atop the LCM and
simplify if possible. Remember the sign in
the original question.
ALGEBRA(Basics 2)
Combining steps 1-3.Simply
2𝑎
𝑏
+
2𝑎𝑑 + 3𝑐𝑏
𝑏𝑑
This cannot be simplified any further.
3𝑐
𝑑
=
ALGEBRA(Basics 2)
Ex.11 Simplify
2𝑗
𝑥
2𝑘
+
𝑦
LCM=xy, After steps 2 and 3 we get…..try it
yourself then go to the next page.
ALGEBRA(Basics 2)
=
Ex.12
3𝑐
𝑑
−
2𝑗𝑦+2𝑘𝑥
𝑥𝑦
2𝑎
𝑏
LCM=?,
Try it yourself before going to the next page.
ALGEBRA(Basics 2)
• Algebra Basics 2 Ends Here go to Algebra
Exercise section and do Algebra Exercise 1
ALGEBRA(Basics 2)
3𝑐𝑏−2𝑎𝑑
=
𝑑𝑏
Remember the sign!
ALGEBRA(Indices)
Key Terms
Index- An index is the power to which a
number is ‘expressed’ that the number of
times it is multiplied by itself.
Eg.
x2=x*x
x3=x*x*x
xn=x*x*x.…n times….x*x*x
ALGEBRA(Indices)
Laws of Indices.
1.am*an= am+n we add indices when multiplying
similar terms. For example a2 * a3 = a5
Because…..
since a2=a*a
and a3=a*a*a
then a2 * a3
=>(a*a)*(a*a*a)
=>a*a*a*a*a
=>a5
ALGEBRA(Indices)
𝑎𝑚 m-n
𝑛 =a
𝑎
2.am/an=am-n or
we subtract indices
when dividing similar terms.
For example a6/ a4 =a6-4 =a2
Because….
a6 = a*a*a*a*a*a
a4 =a*a*a*a
a∗a∗a∗a∗a∗a a∗a 2
6
4
Therefore a / a =
= =a
a∗a∗a∗a
1
ALGEBRA(Indices)
1
-m
3.a =
𝑎
𝑚
For example
1
-2
4 =
4
2
=
1
16
ALGEBRA(Indices)
4.(am)n=amn For example (a2)3 = a6
Because…
If a2=a*a
Then =>(a 2)3
=>(a*a)*(a*a)*(a*a)
=>a*a*a*a*a*a
=>a6
ALGEBRA(Indices)
Laws of Indices. Continued
5.a0=1 For example 30=1. Note any number or variable
raised to the power zero is equal to one.
Ex.1 x2 * x3 = x2+3 = x5
Ex.2 a6 /a3 = a6-3 =a3
Ex.3
2b-2
=
1
2* 2
𝑏
=
2
𝑏2
Ex.4 (2d2)3 = 23*1 * d2*3 = 23 * d6 =8d6 remember deal
with brackets before anything else and a number with
no power is to the power 1.
ALGEBRA(Indices)
Ex.5 This example combines two laws.
𝑏4∗𝑏6
Simplify 2
+
6 4
=
𝑏
𝑏2
=
𝑏
𝑏10
𝑏2
=b10-2
=b8
ALGEBRA(Indices)
Ex.6 If x= 23 * 32 Evaluate x2.
Step 1.Substitute 23 * 32 for x in the terms x2.
(23 * 32 )2. In other words if x= 23 * 32 ,
Then x2=( 23 * 32 )2
Step 2. Evaluate by expanding brackets.
=>23*2 * 32*2
Step 3. Simplify
=>26 * 34
ALGEBRA(Indices)
Ex.6 If x= 23 * 32 Evaluate x2.
Step 1.Substitute 23 * 32 for x in the terms x2.
(23 * 32 )2. In other words if x= 23 * 32 ,
Then x2=( 23 * 32 )2
Step 2. Evaluate by expanding brackets.
=>23*2 * 32*2
Step 3. Simplify
=>26 * 34
ALGEBRA(Indices)
Ex.7 Simplify 3x2(x3+2xy2)
Step 1. Always deal with brackets first
remember BODMAS. Expand brackets.
3x2 *x3+ 3x2 * 2xy2
Step 2.Then simplify.
3x5 + 6x3y2
ALGEBRA(Indices)
Simplify
𝑝2 𝑞 3
𝑝𝑞2
2
Step 1. Expand brackets

𝑝4𝑞6
𝑝𝑞2
Step 2. divide similar index terms(subtract)*
=>p3q4
ALGEBRA(Indices)
Algebra Indices Ends Here go to Algebra
Exercise section and do Algebra Exercise 2
ALGEBRA(Equations and
Inequalities)
An equation is a mathematical statement
which consists of an equal ‘=‘ sign and says
what is on the left hand side of the
equation is equivalent to what is on the right
hand side of the equal sign. For example
4+5=9 or y=x2+2x+3 etc.
ALGEBRA(Equations and
Inequalities)(Key Terms)
DifferenceSum-
ALGEBRA(Equations and
Inequalities)
Ex.1 If x=-1, and y=x2-3x+9 find y.
Simply substitute the value given for ‘x’ into the equation to find
y, writing in the value given for ‘x’ instead of ‘x’
In this case x is given as -1.
So..
y=(-1)2-3(-1)+9
NB: properties of negative numbers
-any negative number by another negative number results in a
positive number so (-1)2 =-1 *-1=+1.
-subtracting a negative number is like adding, so 3(-1)=3*-1
=-3
And -(-3)=+3
y=1+3+9
y=13
ALGEBRA(Equations and
Inequalities)
Equations represents relationships which
cannot be further simplified.
Ex.2 The value of y is the difference
between six times ‘a’ and 3 times ‘b’
squared. Represent this an equation.
y=6a-3b2
ALGEBRA(Equations and
Inequalities)
Equations represents relationships which
cannot be further simplified.
Ex.3 The value of y is the sum of four times
‘a’ and two times ‘b’ squared. Represent this
an equation.
y=4a+2b2
ALGEBRA(Equations and
Inequalities)
Equations can used to find unknowns in situations
where other information is given.
Ex.4 Paul divides an undisclosed amount ($n) of
money among his three daughters Sarah, Melanie
and Paula in a ratio of 2:3:4.The greatest share is
$20 what is the total value Paul gave out($n).
Step.1 Find the value of one share the since greatest
portion is 4 shares=$20, Let ‘s’ represent I share.
$20
4
then 1 s =
= $5.
Because 4s=$20, by dividing both sides of the
equation statement by 4 we get $5 as the value of
one share.
ALGEBRA(Equations and
Inequalities)
Ex.4 continued
Step.2
So if one share is equal to $5
And there where 2+3+4=9(ratio given) shares
Paul share out 9*$5= $40 which is the value
of $n.
ALGEBRA(Equations and
Inequalities)
Ex.5 Aisha has $12 less than Darlene. Kaila has twice as
much as Aisha and Darlene together. If Darlene has ‘$x’,
then Kaila has how much represented as an equation……
Darlene has $x
Aisha has $x-$12 {$12 less than Darlene}
Kaila has 2($x+$x-12){Twice what Darlene and Aisha have
together.}
Then Kaila has 2($2x-$12)= $4x-$24 =$4(x-6){when
factorized}
ALGEBRA(Equations and
Inequalities)(Key Terms)
An inequality a mathematical statement consisting of the following
signs:< less than sign, what is on the left of the sign is greater than what is
to the right of the sign but the value is never equal to the upper
value. For example a<b says a is less than b but never equal to b.
> more than sign, what is on the left of the sign is greater than what
is to the right of the sign but the value ‘b’ is never equal to the upper
value ‘a’. For example a>b says a is greater than b but never equal to b.
≤ less than or equal to sign, what is on the left of the sign is less than
what is to the right of the sign or the value may be equal to the upper
value. For example a ≤ b says a is less than b or may be equal to ‘b’.
≥ less than or equal to sign, what is on the left of the sign is greater
than what is to the right of the sign or the value may be equal to the
lower value. For example a ≥b says a is greater than b or may be
equal to ‘b’.
ALGEBRA(Equations and
Inequalities)
Ex.6 If 1≤x<5 it means ‘x’ may have a value
between 1 and up to 5 but not including 5
since there isn’t a less than and equal sign
before the 5.
Ex.7 If 2<x≤4 this statement means ‘x’ may
have a value greater than two and up to being
equal to 4 but not inclusive of 2.
ALGEBRA(Equations and
Inequalities)
Ex.8 The value of ‘x’ varies between more
than -2 up to but not including 6 represent
this in an inequality statement.
-2≤x<6
Ex.9 The value of ‘x’ varies between more
than -3 but less than 4 what is the range
for these possible values.
-3≤x<4
ALGEBRA(Equations and
Inequalities)
Ex.10 The value ‘x’ varies from greater than
zero up to less than 5, represent this in an
inequality.
0<x<5
Since it is never equivalent to either limit
ALGEBRA(Equations and
Inequalities)
• Algebra Equations and Inequalities Ends
Here go to Algebra Exercise section and
do Algebra Exercise 4
ALGEBRA( Factorizing and
Rearranging)
Remember with all equations what is on the
left hand side of the equal sign is equivalent
to what is on the right hand side of the
equal sign.
RHS=LHS
What ever you do to the right must be
done to the right side to ensure the
equation is true.
ALGEBRA( Factorizing and
Rearranging)
Ex.1
x-2=6 find the value of ‘x’
‘x’ is having 2 subtracted from it to give a
difference of 6. To find the value of ‘x’ do
the opposite to both sides of the equal sign.
That is add to 2 to both sides.
=> x-2+2=6+2
=> x=8
ALGEBRA( Factorizing and
Rearranging)
Ex.2 2a=6, Find the value of ‘a’.
‘a’ is being multiplied by 2 to produce 6, to get the
value of ‘a’ we must do the opposite that is to
divide.
Divide both sides by the coefficient of ‘a’ which is 2.
2𝑎 6
=
2
2
Therefore….
a=3
ALGEBRA( Factorizing and
Rearranging)
Ex.3
6
=3
𝑎
, Find the value of ‘a’
The variable ‘a’ is dividing the number six, as
with the other two examples do the opposite
to both sides. That is multiply both sides by ‘a’
6 𝑎
∗
𝑎 1
3
1
= *
𝑎
1
=>6=3a
Divide both sides by 3
=>2=a
ALGEBRA( Factorizing and
Rearranging)
Rearranging equations is useful to calculate unknown
variables in formulae.
1
V= Ah
3
Ex.4 If
(This the formula for calculating the
volume where A=area and h=height.)
Make ‘h’ the subject of the formula(equation).
1
3
‘h’ has ‘A’ multiplied by it(hence its coefficient) and
multiplied it(another coefficient). Do the opposite to
both sides that is divide. Remember dividing a fraction
is the same as multiplying the inverse.
ALGEBRA( Factorizing and
Rearranging)
1
V= Ah
3
RHS=LHS
Divide both sides by
1
3
which is the same as
multiplying both sides by 3,which is the inverse of
want to get h by itself so divide both sides by A.
1
𝐴
3
1
∗ V=
1
𝐴
3
1
1
3
∗ ∗ Ah
The A and the 1/3 are cancelled out.
3𝑉
=h
𝐴
1
.
3
We
ALGEBRA( Factorizing and
Rearranging)
Ex.5 𝐸 = 𝐼2𝑅, make ‘I’ the subject of the
formula.
again RHS=LHS
I2 is being multiplied by R, to remove R from
this side do the opposite to both sides that
is divide both sides by ‘R’.
𝐸 𝐼2𝑅
=
𝑅
𝑅
The R is cancelled out on the right side.
ALGEBRA( Factorizing and
Rearranging)
𝐸
This gives = 𝐼2
𝑅
But we want I alone, the opposite of squaring a number is
finding the square root.( 𝐼)
So 𝐼2 = I
So we are doing the same to both sides that is square rooting.
𝐸
𝑅
= 𝐼2
Therefore
𝐸
=I
𝑅
ALGEBRA( Factorizing and
Rearranging)
Side note on roots.
A square root is the opposite of raising a value to the power 2 eg. x2 =
x. Just as a value can be raised to the power 2 or 3 or 4 as shown in
‘Indices’ it may be brought down to the root any number of times.
These are called surds
Eg. If x= 3 find x2
x2 =( 3)2 =3
Eg. If x= 3 find x3
Remember x3=x*x*x
So when we substitute x3= 3* 3* 3=
 ( 3* 3)* 3
=> ( 3)2 * 3
=>3 3
ALGEBRA( Factorizing and
Rearranging)
3𝑥+2
Ex.6 If y=
what is ‘x’.
5
As always with equations RHS=LHS.
Multiply both sides by 5 (the opposite)to cancel out the fraction.
3𝑥+2
=
*5
5
y*5
=>5y=3x+2
Subtract 2 (the opposite) from both sides to get 3x by itself.
5y-2=3x+2-2
=>5y-2=3x
Divide both sides by 3 to get x by itself.
5y−2 3𝑥
=
3
3
5y−2
Therefore x=
3
ALGEBRA( Factorizing and
Rearranging)
Ex.7 If y=x2 + 9 what is ‘x’?
The number 9 is added to x2 do the opposite to
both sides to get x2 by itself that is to subtract.
=>y-9=x2+9-9
=>y-9=𝑥2
Now x is squared that means raised to the power 2.
We just want x so we do the opposite and sqaure
root both sides.
=> y−9= x2
=> y−9=x
ALGEBRA( Factorizing and
Rearranging)
Ex.8Make the subject of the equation.
𝑎𝑦
=3
𝑥
=>ay=3x
3𝑥
=>y=
𝑎
𝑎𝑦
=3
𝑥
ALGEBRA( Factorizing and
Rearranging)
Make y the subject of the equation xy-y=4.
Step.1 Factorize y out of the LHS
This can be re-written as (x*y)-(1*y)=4
=> y(x-1)=4
Step.2 Double check by re-expanding the bracket
in your head.
Step.3 Divide both sides by (x-1) to get y by
itself
4
=>y=
𝑥−1
ALGEBRA( Factorizing and
Rearranging)
Factorizing is the process of rewriting mathematical
statements in a shorter form by collecting factors of the
same type.
Ex.9 Factorize the following algebraic expression ax-ay2bx+2by.
The first two terms are multiples of ’a’ hence ‘a’ is a
common factor. They can be rewritten with ‘a’ outside the
bracket giving the original result when expanded.
=>a(x-y)-2bx+2by
With the rest of the statement 2b is multiplies a ‘-x’ and y
hence it is a common factor.
=>a(x-y)+2b(-x+y) hence the statement is factorized and if
re-expanded will give the original statement.
ALGEBRA( Factorizing and
Rearranging)
Eg.10 3mx+6my+3mz
All the terms in this algebraic expression are multiples of 3 and m hence
3m is the common factor of all these terms.
3mx+6my+3mz=(3*m*x)+(6*m*y)+(3*m*z)
Step.1 Write the common factor outside a bracket.
3m()
Step.2 Fill the bracket with terms that when expanded will give back
the original expression.
Remember 3mx+6my+3mz=(3*m*x)+(6*m*y)+(3*m*z)
=(3*m*x)+(3*m*2*y)+(3*m*z)
3m(x+2y+z)
When expanded gives the original expression. Note the coefficient of the
variable ‘y’ was 6m not 3m so we put a 2 to give 6m is we re-expand.