 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Solving for a Parameter in a Linear Equation - Concept

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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The definition of a function can be extended to define the definition of inverse of a function. Along with one to one functions, invertible functions are an important type of function. The definition of inverse says that a function's inverse switches its domain and range. The definition of inverse helps students to understand the unique characteristics of the graphs of invertible functions.

Solving for a parameter, so a parameter is a fancy way of saying a variable. So in this example what we actually want to do is solve for a variable x okay. And this may look a little bit daunting because we have, what 5 letters up there but how we actually do it is very similar to the stuff we've solved before. So say I give you 7x-4=10 we're going to solve this for x. What we'd end up doing is adding the 4 over leaving us with 7x=14 and then divide by 7 leaving us with x=2. this example is pretty straight forward cause we're dealing all with numbers okay. The logic though is exactly the same as what we do over here, okay.
So all these variables just represent numbers okay but instead of giving it to you as 2 times 4 I'm just giving you b times d. So how you actually solve for x is exactly the same as what we did right over there okay. So b times d is just really a number, subtracted over okay. Got a term with x by itself and then we just want to solve for x so just like we did over here where we divided by 7, we just want to divide by the number in front of the x in this case. So divide by a, divide by a, x=c-bd over a, sort of a funny looking answer but it's the only way in this case we can solve for our parameter, solve for our variable x.