Understanding Equations

Equations are all about working with equality. As with our last blog, we will use an example involving money. If you had $50 and then spent $50 dollars you would have $0. Right? Now let's say Bob has some money, he spent $50 dollars and now he no longer has any money. How much money did Bob have? Yes, Bob had $50 dollars. Now let's translate this into an equation, using "B" for the unknown amount of money that Bob has.
B - 50 = 0  
To find out the value of "B" we need to perform mathematical operations while maintaining equality. This is done by performing opposite mathematical operations on opposite sides of the equal sign. For our equation, 50 is subtracted on the left side of the equal sign, so instead of subtracting 50 we can add (opposite of subtraction) on the right side of the equal sign to obtain an equivalent equation.
B = 0 + 50
Which simplifies to tell us;
B = 50
Which of course we already knew from before.

Addition and Subtraction are opposite operations and multiplication and division are opposite operations.

Let's look at another example with Bob. This time Bob has $160 until Friday which is 3 days away. If he buys                                                                                       
a Kindle for $139, how much should he spend on lunch each day for the next 3 days. Let's look at an equation using "L" for lunch.
3L + 139 = 160
So 3 lunches plus the Kindle has to equal to $160. Once Bob purchases the Kindle, that amount is deducted from his total amount of money. So now 3 lunches have to equal $160 - 139.
3L = 160 - 139
Bob now has $21 to purchase lunch for 3 days.
3L = 21
To spend the same amount for each day, Bob would divide the total amount of money by the total number of days.
L = 21/3
L = 7
So Bob has $7 to spend on lunch each day.
Now let's solve a similar equation without a story line.
Solve for 'x'.
5x + 120 = 180
          5x = 180 - 120
          5x = 60
            x = 60/5
            x = 12

The process works exactly the same when it's just letters.
Solve for 'a'
ab + c = d
      ab = d - c
        a = (d - c)/b
Though the process is the same, we can't combine unlike variables.

Even when the problem expands and appears more difficult, the process is still the same.
Solve for 'y'.
8y + 12 = 5y + 6
        8y = 5y + 6 - 12
        8y = 5y - 6
 8y - 5y = -6
        3y = -6
          y = -6/3
          y = -2

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