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Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

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Solving Linear Equations

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Page 1: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y
Page 2: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Special Products• a(x + y + z) = ax + ay + az

• (x + y)(x – y) = x2 – y2

• (x + y)2 = x2 + 2xy +y2

• (x – y)2 = x2 – 2xy +y2

• (x + y + z)2 = x2 + y2 + z2 + 2xy + 2xz + 2yz

• (x + y)3 = x3 + 3x2y + 3xy2 + y3

• (x – y)3 = x3 – 3x2y + 3xy2 – y3

Page 3: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Solving Linear Equations

Page 4: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Isolating the Variable on the Right

2x - 4 = 4x - 10

Page 5: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Isolating the Variable on the Right

2x – 4 = 4x – 10

2x – 4 + 10 = 4x2x + 6 = 4x6 = 2x (½)6 = (½)2x 3 = x

Page 6: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

CHECKING

2x – 4 = 4x – 10 Solution x = 3

2(3) – 4 = 4(3) – 10 ?6 – 4 = 12 – 10 OK

Page 7: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Parentheses

3 – ( r – 1 ) = 2 (r + 1 ) – r

Page 8: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Parentheses

3 – ( r – 1 ) = 2 (r + 1 ) – r

3 – r + 1 = 2r + 2 – r4 – r = r + 24 – r + r = r + 2 + r 4 = 2r + 2 2 = 2r 1 = r

Page 9: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Parentheses

Check

3 – ( r – 1 ) = 2 (r + 1 ) – r 1 = r3 – ( [1] – 1 ) = 2 ([1] + 1 ) – [1] ?3 = 2(2) – 1?3 = 4 – 1 OK

Page 10: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Strategy for Solving Equations

Page 11: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

- x = c

– x = c => x = – c

– x = 2 => x = – 2

Page 12: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Algebra - substitution or evaluation

• Given an algebraic equation, you can substitute real values for the representative values

Perimeter of a rectangle is P = 2L + 2WIf L = 3 and W = 5 then:P = 2 3 + 2 5= 6 + 10= 16

Page 13: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Substitution

• A joiner earns £W for working H hours• Her boss uses the formula W = 5H + 35 to

calculate her wage.• Find her wage if she works for 40 hoursW = 5 40 + 35= 200 + 35= £235

Page 14: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Substitution

• Find the value of 4y - 1 when• y = 1/40• y = 0.5 1• Find F = 5(v + 6) when v = 975

Page 15: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Rearranging formulae• Sometimes it is easier to use a formula if you

rearrange it first• y = 2x + 8• Make x the subject of the formulaSubtract 8 from both sidesy 8 = 2xDivide both sides by 2´2y - 4 = x

Page 16: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Rearranging formulae

• A = 3r2

Make r the subject of the formulaDivide both sides by 3A/3 = r2

Take the square root of both sides A/3 = r

Page 17: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Brackets

• The milkman’s order is 3 loaves of bread, 4 pints of milk and 1 doz. Eggs per week

• Suppose the cost of bread is b, the cost of milk is m and a dozen eggs is e.

• Work out the cost after 5 weeks• = 5(3b + 4m + e)

Page 18: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Brackets

= 5(3b + 4m + e)to ‘remove’ brackets, each term must be

multiplied by 55(3b + 4m + e) = 15b + 20m + 5e• If the number outside the bracket is a

negative, take care: the rules for multiplication of directed numbers must be applied

Page 19: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Brackets

4(3x - 2y)4 3x - 4 2y = 12x - 8y• What about -2(x-3y)?= -2 x - (-2) 3y= -2x + 6y

Page 20: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Factorising

• The opposite of multiplying out brackets• Need to find the common factors• Very important - it enables you to simplify

expressions and hence make it easier to solve them

• A factor is a number which will divide exactly into a given number.

Page 21: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Factorising

• 2x + 6y• 2 is a factor of each term (part) of the

expression and therefore of the whole number.

2x + 6y = 2 x + 2 3 y= 2 (x +3 y)= 2(x + 3y)

Page 22: Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y

Factorising

• 6p + 3q + 9r• 3 is the common factor• = 3(2p + q + 3r)• x2 + xy + 6x• x is the common factor for each term• x(x + y + 6)