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Eliminating fractions. This lesson concerns linear equations with fractions involved in the coefficients and the constants.

I’m assuming that you are already familiar with the content of the fraction lessons in the arithmetic and fractions module.

And if you’re really not very comfortable with fractions, I’d strongly suggest going back and watching all those videos.

They give a thorough treatment of all the operations involving fractions, all the basic rules of fractions there. Here, I’m just gonna do a brief review of a couple key facts.

So fact number one, the product of any fraction and its reciprocal is 1. This means that if a variable is multiplied by a fraction, we can undo this by multiplying by the fraction’s reciprocal.

That’s really important if we want to isolate x, and x is times a fraction, all we have to do is multiply by the reciprocal.

Fraction fact number two, this one doesn’t really have a name, but the basic idea is that if you think about order of operations, multiplication and division are at the same level of priority. We can multiply and divide in any order.

Now I’ll add the caveat, everything in the numerator has to stay in the numerator. Everything in the denominator has to stay in the denominator. If we start multiplying by something we were dividing by, that’s a problem.

But if we just are multiplying and dividing, for example, here we could multiply a times b and then later divide that by c. Or we could start out by dividing b times c and multiplying a by that quotient.

Or we could start by dividing a by c and then multiply b times that quotient. And so we can rearrange into all these forms and all of these are mathematically equivalent.

That’s very important. In particular, this means that we can think of a variable in the numerator of a fraction as the variable times the fraction coefficient.

So 3x/5 is identical to (3/5)x. Those are not two different things. Mathematically they are identical, and that’s very important to realize.

So for example, if we had a very simple equation such as this, well, of course, that 3x/5, we’re gonna rewrite that as (3/5)x. Now we have a fraction times x, and we’re going to multiply this by its reciprocal.

x = (5/3)(2/7). Multiplying through, we get 10/21. This equation simply involved multiplication. Things get a bit trickier when addition or subtraction of fractions is involved. Another equation with fractions. So here we have 7x/6 + 2/3 = 13/2.

Hm. Well, here we will begin by clearing the fractions. You see, we have addition or subtraction and that makes it a little trickier. But one easy way to clear the fractions is to multiply by the least common multiple of all the denominators. Here the denominators are 6, 3 and 2.

Those are the three denominators, and the least common multiple of those is 6. So we will multiply every term by 6. So of course, 7x/6 times 6, the 6s cancel and we just get 7x. 2/3 of 6 is 4. And 13/2 times 6. Well, the 2s cancel and we’re left with 13 times 3, which is 39.

And of course, once we get it into this form with the fractions cleared, then it’s a very easy equation. We subtract 4. We get 7x equals 35, then we divide by 7, and we’re left with x = 5. Here’s another example. For this one we have denominators 2, 4 and 3.

And so the least common multiple of these numbers is 12. So we’re gonna multiply both sides by 12. Each term is gonna be multiplied by 12, and then we’re gonna simplify that.

That just becomes this. Then once we have this we can subtract 4x from both sides, we can subtract 15 from both sides, and then divide by 2 to get x by itself.

x = 3/2. Here’s another practice question. Pause the video, then we’ll talk about this. Okay. So now we’re gonna start by multiplying by the least common multiple, which is 6. Everything gets multiplied by 6.

When we simplify all this, we’re left with 3x + 10 = 8x + 6. Subtract 3x from both sides. Subtract 6 from both sides. And divide both sides by 5. So notice so far we’ve been talking about how to solve equations.

Now we’re gonna go back to expressions for a moment. Another fraction-related skill you may need on the test is simplifying a complex fraction.

A complex fraction is a big fraction with smaller fractions nested in its numerator and denominator. So for example, that is a complex fraction.

To simplify a complex fraction, we use a strategy remarkably similar to the one that we used for equations. Just as we can multiply both sides of an equation by the same number, so we can multiply the numerator and denominator of a fraction by the same number.

To simplify a complex fraction, we multiply the numerator and denominator of the big fraction by the least common multiple of all the denominators of all the little fractions.

So for example, here’s our complex fraction, and we wanna clear the fractions. We wanna simplify this. So the LCM, the least common multiple, of the little denominators is 12. So the little denominators are 2, 4, 3 and 2. The least common multiple is 12.

So we’re gonna multiply the top and bottom of the fraction by 12. When we multiply through, we get 6x plus 15 divided by 4x plus 18.

We can use the same technique to simplify a complex fraction composed solely of numbers, with no variables at all. So for example, if we had this and we wanted to simplify that. Well, the least common multiple of 3, 9, 3 and 2 is 18.

When we multiply by 18, we get 6 + 8 over 12 + 9. That gives us 14/21. We can cancel the 7s, and this equals 2/3. So that whole complex fraction at the beginning simplifies to two-thirds. Here’s a practice problem. Pause the video, and then we’ll talk about this.

Okay. So what we’re gonna do here is multiply by the least common multiple of 2, 6 and 7. That least common multiple is 42. So we’ve multiplied the numerator now by 42, and I’m going to rewrite 42 this way.

I’m gonna rewrite in terms of 6 times 7 or 7 times 6. And the reason I’m gonna do this is that in the numerator, I’m going to keep the 6 outside but multiply the 7 through the parentheses.

In the denominator I’m gonna keep the 7 outside, and multiply the 6 through the parentheses. When I do that I get this.

Well, that polynomial, that quadratic in the numerator x squared minus 10x plus 21, we can factor that to x- 3 and x- 7. Then the (x- 3)’s cancel and we’re left with 6/7(x- 7).

In summary, we can undo multiplication by a fraction by multiplying by the fraction’s reciprocal. When we have multiple fractions added or subtracted, we can clear the fractions by multiplying with the least common multiple of the denominators.

We multiply that on both sides, and we can simplify a complex fraction by multiplying the numerator and denominator of the big fraction by the least common multiple of all the denominators of the little fractions.

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