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Solving Linear Equations - Problem 1
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!
So solving a linear equation with some fractions in here. Most people tend not to like dealing with fractions. The easiest way to solve a problem like this, is to get rid of them all together, which we do by just multiplying by the least common denominator, the LCD.
Here we have a denominator of 3, denominator of 5, they don’t have any common factors, so we basically just have to multiply by their product, 15. Multiply both sides by 15, all of our fraction are going to go away. What people do in this problem is, they forget to distribute. Remember that this 15 has to go to both of these terms, so 15 times 2/3, we have 5, 15 and 3 cancel, so that will give us 5 times 2 this will just be 10x and then 15 times 10, 150.
We could distribute this 1/5 through, but then, we’re just going to have to cancel the with the 15 either way. So what I would do is just say, okay, 15 times the 1/5 is 3, so what turns out to happen is, this goes away, this goes away leaving us with 3 times that. Again, using our laws of distribution, this goes to the x and the 36, 36 times 3 is a little bit bigger. You can use your calculator if you need to, but I believe it is 108.
Just by multiplying by 15, we ended up getting rid of all our fractions. The equation hasn’t changed at all, we just have nicer numbers. From here, same as what we did before, we just want to get our xs together, get our other numbers together. So again to keep our xs positive, I would subtract 3. 10 minus 3 is 7x and then again move this 50 over, so 108 minus 150 is -42 finish it up, divide by 7, x is equal to -6.
Just by multiplying by that LCD we were able to make the equation a lot easier to deal with.
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