We multiply fractions by multiplying numerators and multiplying denominators; and then, of course,
we reduce to lowest terms, if we can, we take the result that we get and reduce to lowest terms. So, a over b times c over d; that is,
the fraction a over b times the fraction c over d, by definition, is a times b and b times d.
That is, the numerator of this product is the product of the numerators;
and the denominator of this product is the product of these denominators.
Multiply straight across the top; multiply straight across the bottom that's the product of two fractions. So, 2/3 times 4/5:
I multiply 2(4) in the numerator, 3(5) in the denominator; I end up with 8/15. That's in lowest terms, so I don't have to reduce.
Even if I have 3/4 times 9, I can think of the number 9 as the fraction 9 over 1: 9 for the numerator, 1 for the denominator.
So, 3/4 times 9 over 1 the numerator will be 3(9) = 27; and the denominator will be 4(1) = 4.
Let's looks at some other examples of multiplication with fractions. Here, I have 9 over 20 times 4/3;
and my instructions say to multiply and write the answers in lowest terms.
So, what I'm going to do is multiply here, 9(4), divide that by 20(3), or put that over 20(3).
Now, I could multiply this out and multiply this out, but then I would just have to reduce to lowest terms, afterwards.
So instead, what I'm going to do is simply factor the factors that I have here: 9 = 3(3); 4 = 2(2);
20 = (4)(5), so 4 = (2)(2) then times 5; and then, that's times 3. So, there's my numerator of the product factored completely;
there's the denominator of this product, factored completely; and I reduce to lowest terms by dividing out factors they have in common.
So, the 3's divide out, those 2's divide out, and those 2's divide out. So, there
I've divided out all the factors common to the numerator and denominator. What's left in the numerator is 3; what's left in the denominator is 5.
And so, the product of 9/20 and 4/3 is the fraction 3/5. I find it by multiplying numerators and multiplying denominators,
but I don't actually multiply them out; I just write them this way, and then, factor the numerators
factor the factors of the numerators; same with the denominators; then, I divide out common factors.
Here's our next example: 3/4 times 12; remember I can think of 12 as 12 over 1, so it will be 3(12).
I'll multiply straight across the top; and then, across the bottom, 4 times 1; reduce to lowest terms.
I'll factor completely: 3(12) = 4(3), so 2(2)(3); and in the denominator, 2(2)(1); divide out factors they have in common;
I end up with 3(3), which is 9, divided by 1, which is, as you know, just 9. So, same process:
I multiply by multiplying numerators, multiplying denominators; then I factor the numerator and denominator completely,
divide out any factors they have in common to reduce to lowest terms.
Problem number 3 involves variables, so I'm going to factor completely. Let's write this as a(a)(b)(c)(c)(c).
So, I've done two steps at once here; I've multiplied the numerators and I've factored them, or written them in factored form, at the same time.
This denominator times this in factored form will be c(a)(b)(b). Now, let's divide out common factors.
The a's divide out; let's see, there's no factor common for that a; so these b's divide out; and then, that c with that c.
So, I'm left with, in the numerator, a(c)(c), which is a(c²); and in the denominator, I'm left with just b.
So, that's the product, in lowest terms, of these two fractions right here. Here's our next example: 8(1/2)² + 9(1/3)² .
So, according to my rule for order of operation, I need to take care of the numbers that have exponents on them first,
and then I'll multiply, and then I'll add. So, (1/2)(1/2) = 1/4 times 8; (1/3)(1/3) = (1/9) times 9.
(1/4)(8) will be 1(8), which is 8, over 4(1), which is 4. This will be 1(9), which is 9 over 9(1). Remember this denominator
and this denominator are both 1's, so 9(1) over 9. So, I end up with 8 over 4 is just 2 when I reduce that to lowest terms;
9 over 9 is 1, and so I end up with just 3, so, another problem that involves multiplication of fractions and that rule for order of operation.
Let's look at our next example: find 1/3 of the sum of 8 and 4. So, what I want to do is translate that into symbols.
I want to find 1/3 of, so (1/3)(8 + 4). So, the sum of 8 and 4 is 8 + 4; I'll add 8 + 4 and get 12.
(1/3)(12); now, if I multiply 1/3 times 12, it'll be 1(12) over 3(1); 12 ÷ 3 then will be simply 4. So, that comes out to be just 4.
Our next problem involves finding the area of a geometric figure.
So, let's take a look at it. We want to find the area. Now, let's see, the dimensions here:
this 12 inches stands for the distance from here all the way over to here. This is 10 inches, and this down here is 20 inches.
So, since this is 20 inches, this must be 20 inches here. Since this is 10, this must be 10, so that tells me that this little thing right here is going to be 2.
So, let's see. I can find, first of all, the area of this rectangle right here. It has a length of 10 and a width of 5;
so that area for that is going to be 5 times 10 (and then, I'll put in my units in a second);
plus, now, the area of this, this distance right here, must also be 10, from here to here, because this is 20; so, this must be 20;
if this is already 10, this must be 10 also; so, now I have, let's see, 8 inches for this length right here, 10 inches for this distance;
so, that area will be 8 times 10. The last thing I have is the area of this little triangle, and from the book,
I know that the area of a triangle is 1/2 times the base, which is 2, times the height, which is 8.
So, let's see, I end up with 50; that's 5(10); plus 80, which is 8(10); plus (1/2)(2) is 1, (1)(8) = 8;
so, I should end up with 50 + 80, which is 130, + 8 = 138 square inches. So, I'll write my units that way on the end: 138 square inches.
So, when you see little figures like this, don't be afraid to draw in some vertical lines or horizontal lines to divide it up
into a number of separate, little simpler figures; find the area of each one of those;
add those areas together that will be the area of the whole figure.