ζYear 8 – Algebraic FractionsDr Frost

Objectives: Be able to add and subtract algebraic fractions.

Starter

Are these algebraic steps correct?

40 - x3

Fail Win!

(Click your answer)

= x + 4 40 3

= 2x + 4

2(4 – 2x) = 3x - 2 2(4) = 5x - 2

Fail Win!√(2 - x) = 2x + 3 √2 = 3x + 2

Fail Win!

Starter

Are these algebraic steps correct?

(x+3)2

Fail Win!

(Click your answer)

x2 + 32

(3x)2

Fail Win!

32x2 9x2

Starter

To cancel or not to cancel, that is the question?

y2 + x2 + x

s(4 + z)s √(x2 + 2) = y + 2

(2x+1)(x – 2)x – 2

pq(r+2) + 1pq

Fail Win! Fail Win!

Fail Win!

Fail Win!

Fail Win!

1 + r2

Fail Win!

- 1

(Click your answer)

Adding algebraic fractions

23+12=46+36=76

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What’s our usual approach for adding fractions?

The same principle can be applied to algebraic fractions.

1𝑥

+2

𝑥2=𝑥𝑥2

+2

𝑥2=𝑥+2𝑥2

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The Wall of Algebraic Fraction Destiny

“To learn the secret ways of the ninja, simplify fraction you must.”

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Exercises

1

2

3

4

5

7

8

9

10

11

6

12

13

14

15

16

17

18

19

20

21

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More Difficult Algebraic Fractions

For most of the examples so far, we’ve considered numerators that don’t contain variables. But when we have variables, the principle is the same!

Lowest Common

Multiple of 3 and 4?

12

Lowest Common

Multiple of x and y?

xy

Lowest Common

Multiple of 2x and 3x2?

6x2

Lowest Common

Multiple of x and x+1?

x(x+1)

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Example

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Step 1: Identify the Lowest Common Multiple.

Step 2: Whatever we multiplied the denominator by, we have to do the same to the numerator.

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Example

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Step 1: Identify the Lowest Common Multiple.

Step 2: Whatever we multiplied the denominator by, we have to do the same to the numerator.

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Examples

2𝑥

+ 3𝑥+1

=2 (𝑥+1 )+3 𝑥  𝑥(𝑥+1)

= 5 𝑥+2𝑥(𝑥+1)

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𝑥+1𝑥−

𝑥𝑥+1

=(𝑥+1 )2−𝑥2

𝑥 (𝑥+1 )= 2 𝑥+1𝑥 (𝑥+1 )

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𝑥−34

+ 𝑥−53

=3 (𝑥−3 )+4 (𝑥−5 )

12=7𝑥−29

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Exercises

1

2

3

4

5

7

8

9

10

11

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( ) =

Multiplying and Dividing Brackets

y2

2x3

× = xy2

6z2

4x3

= 3z2

4x

x3

22 x6

4x+1

3x+2

4 = 4(x+1)

3(x+2)

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Exercises

1

2

3

4

5

7

8

9

10

11

6

y3

2xy× = xy2

2

x2y

xy× = x2

2y2

x+1x2

xy× = x+1

xy

2xy

zq

= 2qxyz

x+1y

z+1q

= q(x+1)y(z+1)

q2

y+1xq

= q3

x(y+1)

( )xy2

= x2

y4

2

( )2q5

z3= 4q10

z6

2

( )3xy

= 9x2

y2

2

( )3x2y3

2z4= 27x6y9

8z12

3

( )x+13y

= (x+1)2

9y2

2

12

( )x+13y

= (x+1)2

9y2

2

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13

14

15

16

18

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17

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Head to Head

vs

Head Table

2

3

4 5

6

7

8 9

10

11

12 13

14

15

Rear Table

Question 1

1

𝑥2+2

𝑥2

Answer:

Question 2

𝑥2

+𝑥4

Answer:

Question 3

23+𝑥+19

Answer:

Question 4

𝑥𝑦

+𝑥+1𝑦 2

Answer:

Question 5

1𝑥

+𝑥𝑦

Answer:

Question 6

1𝑧

+1

𝑧+2

Answer:

Question 7

1𝑥

+1𝑦

+1𝑧

Answer:

Question 8

1𝑥÷3

Answer:

Question 9

2÷1

𝑥2

Answer:

Question 10

1𝑥÷1

𝑥2 𝑦

Answer:

Question 11

( 𝑥𝑦 3 )2

Answer: