Algebra (Period 1)

 LITERAL EQUATIONS AND DIMENSIONAL ANALYSIS 2-8 

We’ve already talked about FORMulas being equations with KNOWN RELATIONSHIPS

The Common Core lingo for a formula is LITERAL EQUATION

Many times, you’ll need to solve for one of the variables.

Example: d = rt
Sometimes, you want to solve for distance, but other times you need the time or the rate.
You can manipulate the variables by balancing the equation until you solve for the wanted variable.
In the formula, d = rt, to solve for rate, divide both sides by t
To solve for time, divide both sides by r
Don't think of this as 3 different equations!
Just learn the main one and use that to solve for what you need!

Example: Solve for l:
P = 2l + 2w
Subtract 2w for both sides: P - 2w = 2l
Divide by 2 on each side:
P - 2w = l
2

When the variable is in the denominator, it's usually easiest to use cross products:
Solve for x:

3z = 4
x      y

Cross products:
3yz = 4x

Divide both sides by 4:
x = 3yz
       4

If you have a fraction, you can simply use the multiplicative inverse:
Solve for b:
3ab = 7c

4          

Either use cross products or multiply each side by 4/3a:
b = 7c (4/3a) = 28c/3a

Why can't you use cross products for the following?
Solve for f:
d = 2a + 2b

      c f