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Solving for a Parameter in a Linear Equation - Problem 2
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!
Solving for a parameter. In this example we’re actually mixing numbers and variables. So we have y, z, x, we also have the number 4 and the number 5. Remember that variables are just representing numbers though. Really, even though it’s a 4 and a 'y' they’re pretty much the same thing. One is a known number, the other is just a symbol for a number, but they’re really just the same idea.
For this one, we’re solving for x, which means we want to get x by itself. Two ways of doing this. We could first split up these two, as fractions, so we could do x over 5 minus 4/5, if you really wanted to, minus 4/5. Just do the properties of distribution, you just split it up. In general, try not to do that, people don’t like dealing with fractions. And so, what we’ve really done is, just made two fractions instead of one. It’s going to make our life a little more complicated.
What you can really do is, try to get rid of your fractions, multiply by the least common denominator. Here it’s 5, here we don’t have any denominator, so 5 is our least common denominator. Multiply both sides by 5, here, our 5 disappears and here our 5 goes to both variables, 5y plus 5z.
In Algebra 1 terms, all we just did is really just cross multiply, but as we move on, we try to sort of use a little bit more accurate, so we’re actually multiplying by the least common denominator. To then solve for x, we know how to do this, it’s just a number, add it over. X is equal to 5y plus 5z plus 4.
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