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Solving Literal Equations: Perimeter of a Rectangle and Two Examples, Summaries of Algebra

The concept of literal equations and formulas, focusing on the perimeter of a rectangle as an example. It also provides two examples of solving literal equations for different variables using algebraic methods. Students will learn how to identify the variable they want to have alone, treat all other letters as numbers, and apply the rules of algebra to solve the equations.

What you will learn

  • What is the formula for the perimeter of a rectangle and how can you solve it for the length or width?
  • What is a literal equation?
  • How do you solve a literal equation for a specific variable?

Typology: Summaries

2021/2022

Uploaded on 09/27/2022

ekani
ekani 🇺🇸

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Literal Equations & Formula
A literal equation is an equation which consists primarily of letters.
Formulas are an example of literal equations.
Each variable in the equation "literally" represents an important part of the whole relationship
expressed by the equation.
For example,
The perimeter of a rectangle is expressed as P = 2L + 2W
The "L" represents the length of one side of the rectangle. We see "2L" in the formula since there
are two sides in a rectangle.
The "W" represents the measure of the width. There is a "width" at opposite ends of the
rectangle, hence "2W".
Solving Literal Equations
To solve a literal equation means to rewrite the equation so a different variable stands alone on
one side of the equals sign. We have to be told for which variable we want to solve.
For example,
If we were asked to solve the equation P = 2W + 2L for W,
we would get W = ( P - 2L ) / 2
We started with " P = ..." and ended with " W = ... ".
That's what it means to "solve a literal equation."
Method:
1. Identify the variable that you want to have alone.
2. Treat all other letters as if they were numbers.
You can add, subtract, or multiply by a letter.
You can also divide by a variable as long as it is never zero.
3. Use all of the rules of algebra that you learned to solve equations in the past.
You may need to add the opposite of a variable to both sides of an equation.
You might have to multiply both sides by the reciprocal of the variable.
All rules of algebra apply.
4. Get the variable you want by itself on one side of the equation.
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Literal Equations & Formula

A literal equation is an equation which consists primarily of letters. Formulas are an example of literal equations. Each variable in the equation "literally" represents an important part of the whole relationship expressed by the equation.

For example,

The perimeter of a rectangle is expressed as P = 2L + 2W

The "L" represents the length of one side of the rectangle. We see "2L" in the formula since there are two sides in a rectangle. The "W" represents the measure of the width. There is a "width" at opposite ends of the rectangle, hence "2W".

Solving Literal Equations

To solve a literal equation means to rewrite the equation so a different variable stands alone on one side of the equals sign. We have to be told for which variable we want to solve.

For example,

If we were asked to solve the equation P = 2W + 2L for W, we would get W = ( P - 2L ) / 2

We started with " P = ..." and ended with " W = ... ". That's what it means to "solve a literal equation."

Method:

  1. Identify the variable that you want to have alone.
  2. Treat all other letters as if they were numbers. You can add, subtract, or multiply by a letter. You can also divide by a variable as long as it is never zero.
  3. Use all of the rules of algebra that you learned to solve equations in the past. You may need to add the opposite of a variable to both sides of an equation. You might have to multiply both sides by the reciprocal of the variable. All rules of algebra apply.
  4. Get the variable you want by itself on one side of the equation.

Example 1:

Solve 5x + 2y = 7 for "y".

Solution:

5x + 2 y = 7 Identify "y" as the variable you want alone. 5x -5x + 2 y = 7 - 5x Add the opposite of 5x to both sides. 2 y = 7 - 5x 2 y = 7 - 5x Divide all terms on both sides by the coefficient of "y" 2 2 2

y = 7 - 5x Here is your solution. 2 2

Rewritten for "y": y = ( 7/2 ) - ( 5/2 )x


Example 2:

Solve T = 2 ππ R(R+h) for h

Solution:

T = 2 πR( R + h ) Identify "h" as the variable you want alone.

T = 2 πR( R + h ) Divide both sides by the factors 2πR. 2 πR 2 πR

T = 2 πR( R + h ) Cancel. 2 πR 2 πR

T = ( R + h ) To solve for h, add the opposite of R to both sides. 2 πR

T - R = R + h - R 2 πR

T - R = h You are done. 2 πR

Hence, h = [ T / ( 2 ππ R ) ] - R