Solving Literal Equations and Formulas
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GOAL: I can solve for a variable in an equation when that variable is in the denominator or a fraction.
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Text reference: Chapter 2, section 5, pages: 109 ─ 114 |
Illinois learning standard: 8.C.3 |
NCTM standard: XX─XX |
Lesson Vocabulary |
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Literal Equation
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A literal equation is an equation that has two or more variables in it. Some examples include:
y = 3x + 2 3x ─ 2y = 10 2x = 7x ─ 16
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Formula |
Formulas are special types of literal equations. A formula is an equation that gives the relationship between the variables. Some examples include:
P = 2•L + 2•W
A = L • W
C = 2 π r |
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Strategies and Content Practice |
Notes |
Begin
(prepare) |
Variables in the Dungeon
Until now, students have never been
asked to solve an algebra problem where the variable they were trying to
solve for (isolate) was in the denominator of a fraction. In
pre-algebra and algebra (up to this point) the variable students seek to
solve for is either in the numerator of a fraction or there is no
fraction at all. Thus students have no prior knowledge solving for
variables in the denominator of a fraction, what I call "variable in the
dungeon". I call them this because the variable is trapped below and
need to be "rescued" from the dungeon (brought up to the numerator)
before they can be isolated and solved for.
For example, take this very simple equation from the 8th grade science, "Motion and Forces" book:

In this equation, the Rate is already solved for; the Distance can be solved for by multiply both sides of the equation by Time; but the Time is stuck in the denominator of the fraction. Even if the Distance could be moved to the other side of the equation, the right side of the equation would still NOT be Time... it would be 1 over Time or
which is not the same as Time.
;)
A Super-Hero algebra tool needs to be introduced and used to rescue Time from the denominator so the equation can be solved for Time.
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Engage |
The Super-Hero tool needed to rescue a variable from the dungeon is multiplication.
In problems like these where the variable being solved for is in the denominator, the rescue technique is to multiply both sides of the equation by the variable you are trying to rescue.
This process accomplishes two things. First, it cancels out the variable in the dungeon, so there no longer is a variable in the dungeon. And second, it moves the target variable to the numerator on the other side of the equation. With the target variable now in the numerator, all that needs to be done is to isolate it and you have the answer.
In the steps above, both sides are multiplied by T to rescue it from the dungeon (denominator). This gives the equation TR = D. To get T all alone by itself, both sides of the equation need to be divided by R. After both sides of the equation are divided by R, T is alone by itself and the problem is solved.
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Assess
(formative) |
Give students selected problems from the practice worksheets below. Circulate to check for understanding. Select students to work their problems at the board and have them explain their answers. |
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Apply |
Practice Assignments.
Worksheet 2-5, Version A
Worksheet 2-5, Version B
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Assess (formative) | Answers to the Practice Assignments. Answers to Worksheet 2-5, Version A
Answers to Worksheet 2-5, Version B
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