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HA1-175 Solving Literal Equations

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1 HA1-175 Solving Literal Equations
Before you do the HA1-175 Lesson. - Watch this powerpoint in slide show mode - Write down the example problems in your notebook. - Do the practice problems in your notebook.

2 What is a Literal Equation
A literal equation is an equation with more than one variable. Examples: A = ½ bh ( area of a triangle) C = 2πr ( circumference of a circle) D= r t ( Distance/rate/time formula) P=4s ( Perimeter) You won’t always know what all the symbols/letters mean, but that won’t stop you from solving for one of the variables using inverse operations.

3 What if I NEED to get ‘s’ by itself? HOW DO I DO THAT??
Let’s take our square perimeter formula P=4s. Sometimes you might HAVE the perimeter, and need to know how many sides there are, so you need to have “s=“ to find that. Perimeter of a square P=4s What if I NEED to get ‘s’ by itself? HOW DO I DO THAT??

4 This process of solving a formula for a given
variable is called "solving literal equations". -The word “literal” is related to the word letters, because there a lot of letters(variables) in these equations.

5 So "solving literal equations" may just be
another way of saying "take an equation with lots of variables, and solving for one of them using inverse operations.”

6 To solve literal equations, you do what
you've done all along to solve equations, use inverses operations List of INVERSE Operations Addition & Subtraction UNDO each other. Multiplication & Division UNDO each other. Squares and Square Roots UNDO each other.

7 except that due to all the variables you won't necessarily be able to simplify your answers as much. The answer may look a little MESSY. Example: Solve D=r*t for “t” The answer is t =D/r The answer is found by moving letters around using inverse operations. Here, each side was divided by variable r.

8 Here's how "solving literal equations" works:
Suppose you wanted to take the formula for the perimeter of a square, P=4s , and solve it for ‘s’ (the side length) instead of using it to solve for perimeter. P=4s How can you get the ‘s’ on a side by itself?

9 Just as when you were solving linear
P=4s Just as when you were solving linear equations, you want to isolate the variable. So, what do you have to do to UNDO the multiplication of ‘4’? The inverse of multiplying is dividing!! You would have to divide EACH SIDE by 4 to isolate the ‘s’.

10 P=4s That’s right, you have to divide by ‘4’. You
also have to remember to divide both sides by 4. LOOK!! We got ‘s’ by itself! This new formula allows us to use the perimeter formula to find the length of the sides of a square if we know the perimeter.

11 Let’s look at another example: Solve this equation for variable d.
Q = c + d 2 2x Q = c + d x2 2Q = c + d -c -c 2Q-c = d Get “d” by itself. We can’t UNDO the addition until we get rid of the 2 under. Multiply both sides by 2, and then CANCEL the 2’s on the right side.        Subtract ‘c’ from each side and then cancel the c’s on the right side. It looks MESSY, but you accomplished your goal and solved for “d”.

12 As you can see, we sometimes must do
more that one step in order to isolate the targeted variable. You just need to follow the same steps that you would use to solve any other ‘Multi-Step Equation’ whether you’re moving variables or numbers. Remember, you MUST UNDO addition/subtraction first, then UNDO mult./Div.

13 Work these on paper before you click to the next page for the answers.
1. d = rt for ‘r’ 2. P = 2L +2W for ‘W’ 3. for ‘t’

14 Check your answers. 1. d = rt for ‘r’ 2. P = 2L + 2W for ‘W’ W = P-2L
Look, each side just had to be divided by “t” to get r by itself. 1. d = rt for ‘r’ 2. P = 2L + 2W for ‘W’ 3. V = 3k for ‘t’ t Vt = 3k Subtract 2L from both sides, then divide by 2 to get w by itself! W = P-2L 2 This one is a little trickier because that “t” is on the bottom. We need to move that first by multiplying both sides by “t” NOW, divide both sides by V to get ‘t’ by itself.

15 But remember, it looks more complicated then it is.
Relax! Just use INVERSE operations to UNDO operations, and you should be fine. These answers may look MESSY, but if you get the variable by itself, it should be right. Now, onto your lesson.


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