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To solve these problems we first look at the fractions. So if we have 1/2 + 1/4. Each fraction has a box of its won. For example here is a link of the fractions and what needs to be shaded in.
You can see that after the two boxes are combined we get 8 small squares. Which results in us shading in 6 squares because of combining all of the shaded in squares.
(1/6)(2/5)
When multiplying fractions you do the same as the above example, you create a box that includes both of the fractions. In the link below is the example of how the boxes should be drawn for a problem like this.
For the first box the blue represents the shaded in part of 1/6. The second box represents 2/5. At the end when we combine the two large squares we see that we end up with 30 small squares, when we combine the shaded portions we see that two squares may overlap. Which is why there are two squares that are shaded in light pink. These two squares represent that the answer is 2/30. That is how you show a multiplying fraction.
In this lesson the topic is based on fractions and how to solve by adding and subtracting. Also to compare two fractions and how to identify which fraction is greater than the other.
Here are a couple examples of fractions that one can compare just by looking at them without doing any math.
The first one is 4/7 and 5/7 we would say this problem as more pieces of the same size. This is because since they have the same denominator then we are able to see which numerator is greater than the other.
Another example would be 3/5 and 6/13 this is Over half way because 6 makes up twice of what 3 is made of.
As to solve problems in addition you have to first make the fractions a common denominator. For instance if you have 32/1 +8/11 you first have to make both of the denominators the same. In this case we will first multiply both 32 and 1 by 11. So that the denominator can be 11. The results would be 352/11 + 8/11 after that we would add the numerators straight across in which we will end up with 360 and the denominator will stay as 11 since they are both the same. The final answer will then be 360/11.
To subtract fractions it would also be the same we would first need to find the common denominator and then subtract the numerators, leaving the denominators the same.
A few tips to know is that the top number also known as the numerator of a fraction represents the number of pieces. The denominator which is at the bottom of the fraction represents the size of the pieces.
Instead of adding, subtracting, and multiplying fractions the way most of us are used to. We can also expand our knowledge by adjusting ourselves to the way our students learn. In the examples below we can solve fractions in a different way, so that some students don't forget to simplify.
Here is an example: 2/3 - 4/9
To proceed we first look at our denominators. From previous lessons we have discovered that a fast way to solve for fractions is by having a common denominator. For this fraction in order to have a common denominator we see that 9 is divisible by 3. Factors that make up 9 are 3x3. So we cross out 9 and replace them with the factors. We look at the first fraction again (2/3) we see that as a denominator it has one 3. To make it equal to the other fraction we multiply it by a 3 and so what we do with the denominator we have to do with the numerator so 2x3=6 and 3x3=9 our new fraction is now 6/9. Knowing that this problem is requiring to subtract we subtract our numerators 6-4 right across and end up with a 2. As for our denominators they are already common denominators so we don't need to do anything to them. We now end up with 2/9 and that is our answer.
Another example that can get students confused is one with fractions that have mixed numbers. Lets say our problem is 10 1/4 - 6 5/8. The first thing that needs to be done is the denominators. We have to convert them into the same number. So lets find the factors of both denominators. So 4 the two factors are 2 and 2. For 8 the two factors that make eight are 4 and 2. As of now you can see that both denominators are not the same. To make them the same we can multiply 2x2 with the number 4 and we also have to do it to the top. So we now have 4/16 for the first fraction. Now the second fraction since we only have 4x2 we have to make the denominator the same as the first fraction, so we can multiply by 2. We also have to do it to the top and we end up with 10/16. When we subtract we now have -6/16. Yes, we end up with a negative. We look at our fractions again and we have to do something about the mixed numbers. 10-6 equals 4. We now have 4 -6/16. That does not seem right. Our answer can't have a negative fraction like that so we cross off our 4 and convert it into a 3, since we are going to be taking away. Our fraction still stays the same -6/16, but in front of that fraction we need a new fraction so that we have something to subtract it by. We look at our denominators once again and since we want an equal denominator we make 16 our new denominator and also as the numerator(what we do at the bottom we do at the top). Our fractions should be like this, 16/16 - 6/16. We now subtract straight across and end up with 10/16. Bringing the whole number, 3, we include it and our final answer is 3 10/16. It took longer than regular equations, but we can't just throw the negative in the front because then students will forget to somplify and will end up with a completely different answer.
In this lesson we learned how to multiply fractions in different forms.
To SHOW how to multiply for fractions we draw a square, the same way we've been doing for adding and subtracting. Lets say we have 3/4(2/3). If you
is how it should look.
The pink boxes represents 3/4. The blue boxes represent 2/3. Since the shading in overlaps in some boxes that is what the orange stands for.
To SOLVE for multiplying fractions
Lets say we have 8/10 x 3/4. The first thing we want to do is simplify every number that can be reduced. So we see here that 8 can be simplified by 2 x 4. 10 can be simplified by 2 x 5. For the second fraction we see that 3 can not be simplified, but we do have a four it can be simplified but its best to leave it the way it is. The reason to that is because we want to cross out numbers that are the same and so 4 can cross out the four in the first fraction.
Every time we cross out a number it is called a FUNKY ONE and so when we cross out the numbers we cross them off with a number 1 over it.
, it should look something like this. Once you are done crossing out the numbers, the numbers that have not been crossed off are now circled. You multiply straight across and that is what our final answer is. Remember that we can't cross off numbers if they are both numerators or denominators.
There are fractions that will have a mixed number. In this case we will be solving with the Backwards C method.
So if we have 2 7/9 and 3 3/10. We will go about it by starting at the bottom and ending at the top. So we multiply 9 x 2 +2 and we get 25. 25 is our numerator for the first fraction now and 9 is still our denominator. For the second fraction we have 10 x 3 + 3= 33 as our new numerator and 10 as our denominator. We simplify every number that can be reduced once again and cross out our funky ones as well. We end up with 55/6. We aren't done just yet. To get our answer we also solve for this by doing upwards division. and our final answer is 9 1/6.
To move forward with the lesson it helped to write down the Divisibility Rules.
2: Even numbers Ex. 72 because it ends with an even number
3: Sum of digits is divided by 3. Ex. 432 >4+3+2=9 (3 goes into 9)
4: Last 2 digits are divided by 4. Ex. 4,312 (Yes, because 4 goes into 12)
5: Any number ending in a 5 or 0. Ex. 24,245
6: Even number and sum of digits divided by 3 (it has to be both can't be one or the other) Ex. 153(NO) Ex. 528(YES)
8: Last 3 digits divided by 8. Ex. 5,240
9: Sum of digits is divided by 9. Ex. 4,374 ->4+3+7+4=18
10: Any number ending in zero. Ex. 22,570
Here are a couple of examples:
1.) 477-Looking at the divisibility rules we go from top to bottom and figure out which one fits into this problem. In this case Divisibility 3 and 9 fits into it.
2.)735- Divisibility rules 3 and 5 fit into it
3.)28,872- Divisibility rules 2,3,4,6,8,9
Prime Factorization
For prime factorization it can be answered with a factor tree or upside down division. Let's say for example that our factor tree is 90, underneath that we would need to find factors that make up 90. In this case it would be 9 and 10. We see now that these two factors can be simplified. What makes 9 is 3 and 3 so we would put that underneath. As for 10 the factors 5 and 2 would be placed under. So as everything has been simplified we now have 3,3,5, and 2. Our prime factorization is 2, and since we see more than one 3 we can make it as 3^2 and 5. So our P.F.= 2, 3^2, 5.
Upside down division
Lets say we have 90 as well. We first start off by setting up the problem. We have to find a number that goes into 90, but that is a prime number and it goes into it evenly. In this case it will be 3. Your problem will now look like 3 L90. To proceed we now see that 3 goes into 90, 30 times so we place that below our (house, which is also the symbol that looks like an L). We see what can go into 30 evenly and it is also 3. We now have 3L30. 3 goes into 30, evenly with the multiple of 10 so we end up with 10. 10 can also be simplified by a prime factor and that is 2. We have 2L10, when simplifying we see that we end up with 5. To know what our prime factors will be we look at every number that is outside of the L. That is 3,3,2,5. So our answer is 2x3^2x5.
Greatest Common Factor
To find the GCF we do all of the previous steps that we have been doing above, but with the addition of comparing the two problems that we have. So lets say after we have found the P.F of two problems we compare them and see what P.F'S they have in common. Lets say we have 60 and 45. For each problem we are going to find the prime factorization, but by doing it with a factor tree or upside down division. Once we do that we see that our prime factors for 60 are 2^2 x 3 x 5. For 45 our prime factors are 3^2 x 5. To find our GCF we only want the numbers that these two problems have in common. That is 3 and 5. We see that both problems have a 3 but with different exponents. Since one is greater than the other we only want the smallest exponent. So our GCF is 3^1 x 5. When we multiply that we get 15 and that is our GCF.
Least Common Multiple
For the LCM we first fine the P.F. Then we gather all of the factors combined together. So using the question from above. With both problems our P.F's are 2^2 x 3 x 5 3^2 x 5. When looking for the LCM we want all of these numbers, but we see that there exponents that are greater than others. Since we have that we want the number with the bigger exponent. That is because multiples are big numbers and we want to end up with the greatest results. So we end up with 2^2 x 3^2 x 5. After that we can multiply it all together. We have 4 x 9 x 5 = 180. Our LCM is 180.
Expanded form
Lets say we have the problem 224 + 437. We would start off by looking at each place value. So we would have 200(since its in the 100's place) + 20 + 4. Underneath that we would then add 400 + 30 + 7. Adding straight down we add 200 +400 which equals 600. Then we would add 20 + 30 = 50. Then 4 + 7= 11. Add those three number (600 + 50 +11) that equals to 661 as our final answer.
Lattice
Lets say we have 26 x 69. Click the link. We multiply 6 x 6 and 6 x 9. We include our answers on the right side of the box. Now on the left side we include the product of 2 x 6 and 2 x 9. That is placed on the left side. There are lines diagonally and that separates each column. So with the columns being separated we add. Just how we see in the link, we add within the link and at the bottom we will see our final answer, 1794.
Base Ten Blocks
Lets say we have 34six. Click the link. The three lines are also known as longs. So we have three logs and we also have 4 circles next to it. The reason why we have three longs is because of the tenths place. Now we see that we have six written at the bottom, that is because that defines how many circles we can make before it becomes a long. If it makes six circles in this case or more than it will turn into a long. So since it didn't the 4 circles remain the way they are.
Array
To find an array we count how many intersection points there are. Click the link. Our problem is 4(5), so we make 5 vertical lines and 4 horizontal lines. The lines overlap and so every time the lines intersect then that is where we pin down a circle and our final answer is a 20. Since that is how many intersections(circles) we have.
Left to Right
Lets say we have 748 + 536. Click the link. So we would first add from left to right in this case. 7 + 5=12. Since it is in the hundreds place it will be placed as 1200. Now we add 4 + 3=7. Since it is in the tenths place we would write our answer as 70. Then we would add 8 + 6 and we would get 14. We should have 1200, with 70 below, and 14 is below that as well. Adding all of that we would end up with 1284.
Area Model
Let's say we have 43(56). Click the link. We would place 40(since its in the tenths place) + 3. Now on the side we would have 50 +6. We would then multiply 40 x 50 which would equal 2000, we would place that inside the first box. Then we would multiply 3 x 50 and would get 150. We do the rest 40 x 6=240 and 6 x 3=18. Add right across we would get 2150 and 258. We now add those two together and 2408.
When showing multiplying groups we can do by separating each number with the base ten blocks. So if we have 13(12) it would look like so. Click the link. The first number is 12, so it will be at the top. It will be written out as 10(since it is in the tenths place) and 2 dots aside (as the one's place). Now for the left side we have to find the number 13, we would write out 10 and the 3 dots underneath. Now when we combine 10 and 10 we get 100, so that would be inside the box. Now we intersect every column so the dot from the top will be multiplied by 10 thats on the left side, so we would drag the 10 inside the box too and we do that to the rest. Now that we have came to the part where the dots will intersect we simply just write dots in those boxes as well, they cancel each other out. Now we add it all up, so we would have 100 + 10 + 10 + 10 + 10 + 10= 156.
Now to solve for a generic rectangle algorithm we would solve by creating a box.The problem is 21(14) Click the link. As you can see we place a 10 above the square + a 4. On the side we place a 20 + 1, representing the 21. We multiply each one of them with each other. So we would start by multiplying 10 x 20= 200. 4 x 20=80. 10 x 1=10 and 4 x 1=4. We then add straight across and end up with 280 and 14. Add that all together and we get 294.
Scratch method
Lets say we have 23six + 14six + 31six + 25six + 32six. Click the link. We line up the numbers and since we have the base six every time we get to six or go over the number six we cross out the number. Instead we now write the difference. The first number we would cross is 4, because 3+4=7 and that passes over 6. We figure out the difference of how 7-6 and our answer is 1, so we would replace the 4 with a 1. We do that for the rest of the numbers. When we get to the bottom the last number that was crossed is 5 and it is replaced with a 1. We now add the last numbers and that number goes into the final answer as the (ones place). We do the same for the rest of the problem and then we get out answer 213six.
Expanded form
Lets say we have the problem 21five + 13five. We would start off by making 20 + 1. The reason for the 20 is because(its in the tenths place). Underneath that we would put 10 + 3. Adding straight down we would have 30 + 4. Add straight across and we would have 34 make sure to include the base so we would now have 34five.
Left to Right
Lets say we were to have 752 + 431. We would solve this by starting from left to right. We see that the number at the left is a 7 and also a 4. We add them together and 1100 would be our answer. Then we have 5 and 3, we would then place 80 underneath and 2 + 1= 3 so that would be our last number below. Each number represents the place in which they are in like the hundredths, tenths, and ones place. We would then add 1100, 80, and 3 straight down and we would get 1284.
Trading off
Lets say our problem is 193 and 107. It is important to know that when we do one thing to a number we also have to do it with the other number as well. To trade off numbers it can be easier to convert the 193 to a 190 and the 107 to a 110. We see there how trading off each others numbers we got whole numbers.
Friendly numbers
The problem is 28 + 62. It is easier to convert the numbers into a tens factor, so ending it with a zero. In this case we would make 28 convert to a 30 and 62 would convert into a 60. Add those together and we get 90 as our final answer.
Converting
If we have the problem 3,461seven. We would start off by writing the amount of numbers there are on top. Click the link. In this case it can count as an exponent. In parenthesis we include (3 x 27) + (4 x 16) + (6 x 6) + (1 x 1)= 81 + 64 + 36 + 1= 182. Make sure to include the base, but since we are converting to base ten then our base would be ten. We would have 182ten.
Converting bases
The problem would be 36four. The first thing we would do is draw the question like in the link. So we have to SHOW the equation. We would first start off by drawing three longs, we have to now draw circles that represent the 6, but we cant forget about our base. So we begin to draw circles and we stop right at four because of our base and we still need to make 6 circles in total. So the first four circles turn into a long and when only include two circles after that. We now have 4 longs and 2 circles. Now to solve we would go about it by writing a 3(4) + 6(4). We basically multiply each number by the base. After that we would get 12 + 24 and that would equal 36.
Converting bases
Lets say we have 73 eleven. We would go about it by first drawing 7 longs and 3 small squares(circles) representing the 3. Since our base is eleven each of the long counts as 11, so with there being seven longs it would add up to 77. So 77 + 3= 80 and that's how we would convert to base 10.
Converting to other bases
Lets say we have 49 to base five. The first thing we would do is is draw 4 longs and we would also draw 9 circles. We look at the base and see that it can only go up to six, once it reached 6 or more that makes a new long. In this case we would have 5 longs once it reaches over the base and we would have 4 circles. It's important to also know that once there are five longs then that turns into a flat. So we see that we now have one flat, no longs(since they have been converted to a flat), and we have 4 circles. So our answer is 104.
There is a lot to cover during this lesson. The first thing will be expanded form. Lets say we have the problem 27(36). In expanded form we will have to literally expand the numbers. So we would start off by making a 20 + 7 at the bottom of that we will multiply by 30 + 6. Click the link, you will then see how we crossed multiplied every number and ended up with 600 + 210 and 120 + 42. Solving for that straight down we would then get 972.
Now lets move forward to an Array. For the array we will be using lines and points. So if we have the problem 3(4) we would have 3 lines facing one way and 4 lines facing the other way. Click the link to see how. So as you can see we drew 3 lines horizontally and 4 lines vertically, they are also intersecting. Now we count how many times the lines intersect with one another. As you can see that is why the circles are there. They represent where the lines intersect. Now we add all of the circles and that is our final answer. Which is 12.
Now we move on to Lattice. Click the link. Now that picture in the link is already solved. So now I will be explaining it. We create a box and every time we see a number we make a line diagonally. So we see that the left side numbers of the box are a different color and that is because they represent the 2 x 1 being multiplied and the 2 x 5 being multiplied. As for the the right side numbers it represents the 5 x 1 and the 5 x 5. Since we now have the sum of every number we then look at where the lines separate the boxes. So we add up all the numbers that are in each like and thats how we end up with 375 as our answer.
To do upwards division we would solve by setting up the problem as a fraction. So the first number will be as the numerator and the second number as the denominator. So lets say we have the problem 69/6. Click the link. We will be looking at it as if we are dividing regularly with the box inside. So back and forth we will be looking at the numerator and denominator and making sure that each number goes into it. So 6 and 6 go into each other 1 time we write the one above the numerator. 6 goes into 9 one time (by 6) so we write the 6 above the 9. We see that there can be a number left over which is 3 when you subtract 9-6. Now we have our answer as 11 3/6. The 11 represents the numbers we found that go into one another and that was one and one which led to 11. 3/6 is the number left over(3) and 6 is the denominator we originally had.
This can be helpful because it can be a quicker way to solve to get an answer than what we normally do.
Lastly, we will be discussing Repeated Subtraction. In a sense this is like division, but it can be a but different. Lets say we were to have 142/8. We would set the problem up by doing it the traditional way with the first number inside the (house/box). Click the link. We see in the link above that how we simply just divided and subtraction until there was no number left to reduce. Instead of traditionally showing all of the work above the box we showed it at the bottom. For every time that we divide like 8 x 9 = 72 we would place the other factor so (9) on the right side next to the line and we would do that for every time we found a factor that can possibly go into the number. We then add up all the numbers on the right side of the line so in this case it would be 17. On the left side at the bottom we had left over a 6 and at the top the number that is outside of the bottom is an 8. So our answer is 17 6/8.
Example: 4+(-7) In this problem we are going to be SHOWING how to solve for it.
We would go about this problem by looking at the first number which is 4. We can see that it is a positive, so underneath we would draw ++++(four positive symbols). The next thing we do is look at the number inside the parenthesis and that is -7. Underneath the positive symbols we are then going to draw -------(seven negative signs). It should now look like the positives are above the negatives and are paired up. So we then circle the symbols that have a pair and circle them. Look to see what are the remaining symbols we have only negatives as left overs, there are three of them. So the answer is -3.
Now that we have seen an example from the notes we can take into consideration that, when the signs are different we subtract. Also keep in mind about the Keep, Change, Change rule. For the first number you always keep , Change the symbol so either positive or negative, and Change the number either to e negative integer or positive.
So for instance if we had the problem 42+(-50). We would first approach this problem by writing the signs above. So 42 seems to be a positive so we write + above it. Then we look at what is inside the parenthesis it's a negative number so we put a - above it and since the bigger number is a negative we solve this problem by subtracting. We subtract 50-42 and we end up with 8. Don't forget to bring over the negative sign since that's what our bigger number represented.
