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8/11/2019 Chapter 11 Algebra 2 2014
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Form 2 [CHAPTER 11: ALGEBRA 2 ]
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11 .1 Solving equations Part 1Example 1: Solve the equation: x + 5 = 12
x + 5 = 12 x = 12 5 x = 7
Example 2: x + 9 = 11
Example 3 : x 10 = 0
Example 4 : 9 + x = 16
Example 5 : 3 x = 1
Check!7 + 5 = 12Correct!
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11 .2 Solving equations part 2 Example 1: Solve the equation 5 x = 20
Example 2: Solve the equation 2 x = 6
Example 3: Solve the equation3
x = 5
Example 4: Solve the equation5
x
= 4
Example 5 : Solve the equation7
x
= 3
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11 .3 Solving equations part 3Example 1: Solve:
3 x + 4 = 16
Step 1: Start by removing the number from the side where there is the unknown.
3 x + 4 = 163 x = 16 4 (Subtract 4 on both sides)3 x = 12
Step 2: We must remove the coefficient of x (in this case 3)
3 x = 123 x = 12 3 (Divide by 3 on both sides) x = 4
Example 2 : Solve the equation 2 x 4 = 10
Example 3: Solve the equation 9 3 x = -33
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Example 4: Solve the equation 4 x + 2 = 14
Example 5: Solve the equation 20 6 x = 2
11 .4 ! Solving Equations part 4. Example 1: 2 x + 6 = 3 x
Step 1: Get all x s on one side of the equation (try and keep the x positive)2 x + x + 6 = 3 (Add x on both sides)
3 x + 6 = 3 (Collect like terms)
Step 2: Solve simply by getting x subject of the formula
3 x + 6 = 33 x = 3 6 (Subtract 6 on both sides) 3 x = ! 3 (Collect like terms) 3 x = ! 3 3 (Divide by 3 on both sides) x = ! 1 (We must be very careful for the SIGNS)
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Example 2Solve the equation 5 x 7 = 5 x
Example 3
Solve the equation 2 x + 21 = 8 x + 3
Example 4Solve the equation 9 + x = 4 4 x
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11 .5 ! Solving equations part 5. Example 1
2 x + 3 x + 5 = 3 x + 4 x 6
Step 1 : We must first ALWAYS collect like terms
x + 8 = 7 x 6 (This is a recognized type of equation which can be worked out normally)
Step 2 : Get numbers on one side and letters on the other
8 = 7 x x 6 (Subtract x on both sides)
8 = 6 x 6
8 + 6 = 6 x 6 (Add 6 on both sides)
14 = 6 x
14
6 x= (Divide both sides by 6)
!
!! !
Example 2
Solve the equation 3 x + 2 + 2 x = 7
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Example 3
Solve the equation 1 4 3 + 2 x = 3 x
Example 4
Solve the equation 9 + 5 = 3x + 4x
Example 5
Solve the equation 4 x 2 x = x
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11 .6 ! Solving equations with brackets. Example 1
Solve the equation 5 ( x 3) = 35
5 ( x 3) = 35
5 x 15 = 35 (Expand the brackets)
5 x = 35 + 15 (Add 15 on both sides)
5 x = 50
5 x = 50 5 (Divide both sides by 5)
x = 10
Example 2
Solve the equation 3( x + 4) = 24
Example 3
Solve the equation 5(2 x + 1) = 4( x 2) + 10
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Example 4
Solve the equation 2( x + 4) = 3(2 x + 1)
11 .7 ! Solving equations with cross multiplication. Example 1
Solve the equation !!
! ! ! !
4 92
y ! =
We must ALWAYS start by removing the number from near the unknown.
9 42
y= + (Add 4 on both sides)
132
y=
To remove the denominator we know that y is divided by 2 therefore to remove that we
must multiply by 2 on both sides.
y = 13 " 2 (Multiply by 2 on both sides)
y = 26
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Example 2
Solve the equation3
2 105
x ! =
Example 3
Solve the equation 1 6 83
x + =
Example 4
Solve the equation 4 102
x ! =
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Example 5
Solve the equation2 6
4 2
x x+ !=
11 .9 ! Constructing Equations.Steps for setting up equations
Read the problem Assign variables Make a list of known facts, translate them into mathematical expressions. Sketch the
problem if possible.
Solve the equation
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Example 1
The perimeter of a rectangle is 48 cm. Its length is x + 8 cm and its width is x cm. Find the
value of x.
The length is 8cm more than the width.
The perimeter of the rectangle (Add all the lengths)
x + 8 + x + x + 8 + x
4 x + 16
This perimeter is equal to 48cm
4 x + 16 = 48
4 x + 16 = 48 16 (Subtract 16 on both sides)
4 x = 32
4 x = 32 4 (Divide by 4 on both sides)
x = 8cm
Length = x + 8 = 8 + 8 = 16cm
Width = x = 8 cm
x + 8
x
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Example 2
I think of a number. When I triple it and add 4 we get the same answer as when I multiply
the number by two and add 6.
Example 3
I think of a number. When I double it and subtract it from 10, the result is 4.
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Example 4
Gilda thinks of a number and adds 7 to it. She then multiplies her answer by 4 and gets
64. What was her original number?
Example 5
The areas of these two shapes are equal.
Find the value of x .
x
15
33
2 x
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12 .10 ! Substituting.Substitution is understood from the meaning of the word. We are going to be given a formula
with a number of variables. We shall also be given the value of the variables. These values shall
be substituted instead or the corresponding letters.
The formula for the area of a rectangle is A = lb
If a rectangle is 3cm long and 2cm wide, we can substitute the number 3 for l and the number 2
for b to give:
l = 3cm and b = 2cm
A = lb
A = l x b
A = 3 x 2
A = 6cm 2
When we substitute numbers into formulas we may have a mixture of operations:
i.e. ( ), x, , + , !
Remember to use the BIDMAS rules whilst working the value of the formula.
Example 1
N = T + G, find N when T = 4 and G = 6
N = T + G
N = 4 + 6 N = 10
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Example 2
If P = 2 (l + b), find P when l = 6 and b = 9
Example 3
If C = RT, find C when R = 4 and T = -3
Example 4
Ifa b
Dc
!= , find D when a = ! 4, b = ! 8 and c = 2
Example 5
If y = m x + c, find y when m = 4, x = ! 2 and c = ! 3
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Example 2
Make s subject of the formula in the equation n = m 3s
Example 3
Make r the subject of the formula for the equationq r
p s
+=
Example 4
Make b subject of the formula for the equation s = 3(a + b)
Example 5
Make h subject of the formula for the equation( )
2
h a b A
+=