Symbolic Exploration Day 3 Module 15 Level IX

Symbolic Exploration Day 3 (Module 15) Level IX of IX Distribution Warm Up Find the Number Guided Practice Distributing the Negative Sign Distributing a Coefficient Distributing a Fraction Anything New

Find the Number (Warm up) Use combining like terms to solve the following problems. 1) – 2 x – 3 + 6 x – 4 +2 = 6 – 2 x + 4 x – 3 x=4 2) What are the measurements of the rectangle below if the perimeter is 20 cm? n 2 n+1 Since n=3, then the length is 3 cm and the width is 7 cm.

Distribution (Guided Practice) Distributing the Negative Sign (Tiles) 1 – Notice there is a negative sign outside the parentheses. What do you think it means? The negative sign means “OPPOSITE”.

Distribution (Guided Practice) Distributing the Negative Sign (Tiles and Abstract) 1 - –(– x + 2) = – 6 x– 2=– 6 x– 2+2=– 6+2 x=– 4

Distribution (Guided Practice) Distributing the Negative Sign (Abstract) 2 – (– x + 3) = 2 x– 3=2 x– 3+3=2+3 x=5

Distribution (Guided Practice) Distributing a Coefficient (Tiles) 3 3 Notice there is a coefficient of 3 outside the parentheses. What do you think it means? Every tile inside the parenthesis is going to increase by a factor of 3.

Distribution (Guided Practice) Distributing a Coefficient (Tiles and Abstract) 3 3 3(2 x+4) = – 6 6 x + 12 – 12 = – 6 – 12 6 x = – 18 x = – 3

Distribution (Guided Practice) Distributing a Coefficient (Abstract) 4 3(x + 2) = x + 2 3 x + 6 = x + 2 3 x – x + 6 = x – x + 2 2 x + 6 = 2 2 x + 6 – 6 = 2 – 6 2 x = – 4 x = – 2

Distribution (Guided Practice) Distributing a Negative Coefficient (Tiles) 5 – 2 Notice there is a coefficient of – 2 outside the parentheses. What do you think it means? Every tile inside the parentheses is going to increase by a factor of 2. Then you will flip the tiles to represent the “opposite”.

Distribution (Guided Practice) 6 – 2 Distributing a Negative Coefficient (Tiles and Abstract) – 12 = – 2(– 2 x + 4) – 12 = 4 x – 8 – 12 + 8 = 4 x – 8 + 8 – 4 = 4 x – 1=x

Distribution (Guided Practice) Distributing a Negative Coefficient (Abstract) 7 – 2(x + 3) = 2(x + 1) – 2 x – 6 = 2 x + 2 – 2 x – 6 + 2 x = 2 x + 2 x – 6 = 4 x + 2 – 6 – 2 = 4 x + 2 – 8 = 4 x – 2 = x

Distribution (Guided Practice) Distributing a Fraction (Tiles) 8 1 _ 2 Notice there is a coefficient of ½ outside the parentheses. What do you think it means? You will have to decrease the number of tiles by a factor of ½.

Distribution (Guided Practice) 9 _ – 1 2 Distributing a Fraction (Tiles and Abstract) – 12 = (– 2 x + 4) – 12 = 1 x – 2 – 12 + 2 = x – 2 + 2 – 10 = x

Distribution (Guided Practice) Distributing a Fraction (Tiles) 10 1 _ 2 Again, notice there is a coefficient of ½ outside the parentheses. Is it easier to solve equations with fractional coefficients abstractly or with tiles? Let’s see… It’s easier to manipulate these type of equations abstractly.

Distribution (Guided Practice) Distributing a Fraction (Abstract) 11 ½(x + 4) = 8 ½(x) + ½(4) = 8 ½x + 2 – 2 = 8 – 2 ½x = 6 2(½x) = 2(6) x = 12

Distribution (Guided Practice) Anything New? 12 6 ?

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