In this lesson you will solve equations without explicitly using bags and marbles to help you. It is important, though, that you will still use the strategies you used when solving the balance puzzles. When solving equations (without the use of the bags and marbles model) you need to understand that the strategy remains the same: get the variables on one side of the equation and the constants on the other. 
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Example 1 

Here is the given equation. Students can first change it, so that it involves only addition if they wish. 

Remembering that our strategy is to change the puzzles so that the bags are on one side and the marbles on the other, we will begin by eliminating the variable on the left side of the equation. 

Now that there are variables only on the right side of the equation, we need to eliminate the constants on the right side. We can accomplish this by adding 9 to both sides. 

The equation "12 = 3n" means there are 12 marble on the left side of the balance and 3 bags on the right. Another way to think of this equation is "Three times what number equals 12?" Dividing both sides by 3 will give us the value of n. 

The value of n is 4. We can verify this solution by substituting it into the original equation. $\matrix{4n + 3 &=& 7n – 9\\4(4) + 3 &=& 7(4)  9\\16 + 3 &=& 28  9\\19 &=& 19}$ 
Example
Find the value of p: $6p4=2p+8$ 
Solve it. $\matrix{6p4&=&2p+8\\+4&&+4\\&&\\6p&=&2p+12\\2p&&2p\\&&\\4p&=&12\\\div~4&&\div~4\\&&\\p&=&3}$ 
Check it. $\matrix{6p4&=&2p+8\\(6)(3)4&=&(2)(3)+8\\184&=&6+8\\14&=&14\\&\large~\surd&}$ 
Example
Find the value of e: $153e=6e3$ 
Solve it. $\matrix{153e&=&6e3\\+3&&+3\\&&\\183e&=&6e\\+3e&&+3e\\&&\\18&=&9e\\\div~9&&\div~9\\&&\\2&=&e}$ 
Check it. $\matrix{153e&=&6e3\\15(3)(2)&=&(6)(2)3\\156&=&123\\9&=&9\\&\large~\surd&}$ 
Example
Lines CD and AB intersect at point E. Angle AEC and Angle DEB are vertical angles. Find the value for m. 
Solve it. $\matrix{2m+10&=&5m80\\+80&&+80\\&&\\2m+90&=&5m\\2m&&2m\\&&\\90&=&3m\\\div~3&&\div~3\\&&\\30&=&m}$ 
Check it. $\matrix{2m+10&=&5m80\\(2)(30)+10&=&(5)(30)80\\60+10&=&15080\\70&=&70\\&\large~\surd&}$ 
Question 1 Find the value of the variable. 7a + 12 = 36 + 4a

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Question 2 Find the value of the variable. 3e + 15 = 6e – 3

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Question 3 Find the value of the variable. 5r + 8 = 15 – 2r

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