chapter 11 algebra 2 2014

8/11/2019 Chapter 11 Algebra 2 2014

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Form 2 [CHAPTER 11: ALGEBRA 2 ]

1 C.Camenzuli| Stella Maris, College

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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

+=

chapter 11 algebra 2 2014

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