Unit 2 - Literal Equations Flashcards

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An equation containing more than one variable

False (B)

y = (x + a)/b

c = 5 - ab Signup and view all the answers

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

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

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

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

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

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

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

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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} ).