Unit 2 - Literal Equations Flashcards
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Unit 2 - Literal Equations Flashcards

Created by
@GrandDwarf5939

Questions and Answers

What is a literal equation?

An equation containing more than one variable

A literal equation has only one variable.

False

Solve for y in the equation x + a = yb.

y = (x + a)/b

Solve for c in the equation ab + c = 5.

<p>c = 5 - ab</p> Signup and view all the answers

Solve for y in the equation xy + p = 5.

<p>y = (5 - p)/x</p> Signup and view all the answers

Solve for b in the equation ab + c = d.

<p>b = (d - c)/a</p> Signup and view all the answers

Solve for a in the equation ab + c = d.

<p>a = (d - c)/b</p> Signup and view all the answers

Solve for b in the equation 3a + 2b = c.

<p>b = (c - 3a)/2</p> Signup and view all the answers

Solve for c in the equation 3abc + b = 5.

<p>c = (5 - b)/(3ab)</p> Signup and view all the answers

Solve for p in the equation i = prt.

<p>p = i/(rt)</p> Signup and view all the answers

Solve for L in the equation P = 2L + 2W.

<p>L = (P - 2W)/2</p> Signup and view all the answers

Study Notes

Literal Equations Overview

  • A literal equation involves two or more variables.
  • It is incorrect to say a literal equation has only one variable.

Solving Literal Equations

  • For the equation ( x + a = yb ), solving for y results in ( y = \frac{x + a}{b} ).
  • The equation ( ab + c = 5 ) can be solved for c as ( c = 5 - ab ).
  • When solving ( xy + p = 5 ) for y, the solution is ( y = \frac{5 - p}{x} ).
  • The equation ( ab + c = d ) yields ( b = \frac{d - c}{a} ) when solved for b.
  • To solve ( ab + c = d ) for a, the equation becomes ( a = \frac{d - c}{b} ).
  • In the equation ( 3a + 2b = c ), solving for b gives ( b = \frac{c - 3a}{2} ).
  • Solving ( 3abc + b = 5 ) for c results in ( c = \frac{5 - b}{3ab} ).
  • For the equation ( i = prt ), solving for p leads to ( p = \frac{i}{rt} ).
  • In the perimeter equation ( P = 2L + 2W ), solving for L results in ( L = \frac{P - 2W}{2} ).

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Test your knowledge of literal equations with these flashcards. Each card presents a definition, identifies a key concept, or provides a problem to solve for a variable. Strengthen your understanding of how to manipulate and solve equations with multiple variables.

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