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GED Math Test Tips [Video Transcript]
The first thing we’re going to do is order numbers from least to greatest on a number line. We’re ordering -2, 5/6, 0/9, 1.2, -1 and 1/5. So first -2, well that’s already on our number line. So we’re going to put a dot there and just to remind yourselves we’ll say -2, 5/6, that’s going to be a little bit more than 1/2 and a little bit less than 1. So 5/6 will be somewhere around here, 0.9 will also be a little bit less than 1, so we need to figure out between these 2 what comes first, and we’ll do that in a little bit, 1.2 will be right about, here and -1 and 1/5, here’s -1 and 1/5 will be a little bit further, so we have -1 and 1/5.
So now we need to determine between 5/6 and 0.9, which one comes first. To do that, it might be best to change 5/6 into a decimal, so we say 5 divided by 6, can’t do that. So we put decimal 0, put our decimal right at the top, 6 goes into 5 8 times, which is 48. Subtract and get 2, bring down another 0, 6 goes into 2 3 times to be 18, 2 to bring down another 0 and that’s just going to repeat. So comparing 0.9 and 0 point 8 3 repeating, this would come first, which means 5/6 is smaller than 0.9. So if we put them in order from least to greatest it’ll be -2, -1 and 1/5, 5/6, 0.9, and 1.2.
Now we’re going to work with multiples and factors. Factors are numbers that divide evenly into a given number. So if you’re looking at factors of 12, you want to find all the numbers that go into 12 with no remainder. You can always start with 1 and whatever number you’re talking about, so 1 and 12. Two goes into 12 and that’s paired up with 6 and then 3 is paired up with 4. So the factors of 12 are 1, 2, 3, 4, 6, and 12. Multiples are the product of a given number and any nonzero whole number, so it’s like we’re going to skip count by 12. So 12 will be the first multiple of 12, and then 12 times 2 is 24, 12 times 3 is 36, 12 times 4 is 48, and then 60, and then it keeps going on and on forever. So notice there’s a set number of factors but there’s an infinite number of multiples.
So some things we can do with factors and multiples are find the greatest common factor, which is called GCF, or the least common multiple, LCM of two numbers. Let’s try to find the GCF and the LCM of 20 and 36. We can find these in two different ways. The first method is to list. So we are going to write down 20 and 36 and we’re going to start by listing the factors for both of them. For 20 it will be 1 and 20, 2 and 10, 3 doesn’t go in but there’s 4 and 5. For 36 it will be 1 and 36, 2 and 18, 3 and 12, 4 and 9, and then 6 times 6, so we only write it once. So if you’re finding common factors, well, they both have 1, they both have 2, they both have 4. So the greatest common factor, greatest means biggest common factor, that will be 4. So you would say the GCF of 20 and 36 is 4.
So let’s do multiples now. So the multiples of 20 will be 20, 40, 60, 80, 100 and so on. We’ll stop there for now, 36 will be 36 times 2 is 72, times 3 is 108, times 4 will be 144, and times 5 will be 180. That could go on and on also. We see here that one of the multiples of 36 is 180, all the multiples of 20 are either going to end in 20, 40, 60, 80, or 00. So the next multiple here that would match up with something on this list would be 180. So our least common multiple of 20 and 36 is 180. Now you can also find GCF and LCM by prime factoring. When we prime factor we want all of our numbers to be prime numbers. People often use factor trees to prime factor, but we can do it without one. So we’re going to say 20 is 2, because that’s where our smallest prime factor, we’ll start with 2, 2 times 10. Ten can be broken up into 2 times 5. For 36 it’ll be 2 times 18, 18 will be 2 times 9 and 9 is 3 times 3.
For GCF what we’re going to do is find our common factors and circle them. So we’ve a pair of 2s and another pair of 2s, so we’re going to build our GCF. We’re going to take one number from each pair and write it down and then we multiply them together, so our GCF is 4, which we found before. For LCM you start the same way, your prime factor, you circle your common factors, you write your common factors down, one from each pair, but then you want to take all the extra numbers. So 3 times 3 times 5, and now you’re going to multiply all of those together. Two times 5 gives you 10, and then 2 times 3 is 6, 6 times 3 is 18, 10 times 18 is 180. Let’s use the distributive property to multiply 4 times 107. You can certainly sit there and do long multiplication, but we’re going to split the number up so that’s easier to do mentally. So we’re going to say 4 times 107 is the same thing as 4, inside the parentheses we’re going to split 107 up. A nice number that’s close to 107 is 100, but to make sure we don’t change the value of our parentheses, we’re going to say 100 plus 7. So this still equals 107 and we’re still multiplying it by 4.
Now the distributive property says if you have a number on the outside of parentheses and on the inside we’ve plus or minus, we can take the number on the outside and multiply it by both things on the inside. So that will mean this is 4 times 100, we bring down our plus and then 4 times 7. So we get 400 plus 28 and our answer is 428.
Here we’re going to apply rules of exponents in numerical expressions with rational exponents to write equivalent expressions with the rational exponents. The first thing is when you’re multiplying the same base you can add the exponents together, so we have 4 to the 5th times 4 and to the 7th, our base is 4 and our exponents are 5 and 7. So we’re going to say 4 and our exponent, we’re going to add them together, and 5 plus 7 is 12, so it’s 4 to the 12th power. When we’re dividing the same base we can subtract the exponents. So we’ve 7 to the 7th divide by 7 to the 3rd, our base stays the same, but now we subtract our exponent, so our answer is 7 to the 4th.
Anything besides 0 raised to the power of 0 equals 1, so -6 raised to the 0th power is 1. When you raise an existing power to another power you multiply the exponents together. So here we have 3 to the 5th but then we’re raising that to the 6th power. So we’re going to say 3 and multiply our exponents together, which equals 3 to the 30th. Now a number raised to a negative exponent means you send it to the denominator of a fraction, so 3 to the negative second power means we’re going to create a fraction and it’s going to go into the denominator, but when you put in the denominator your negative exponent becomes positive. And we can’t just leave a blank in the numerator, so we fill it in with a 1. So it’ll equal 1 over 3 squared or if we want to simplify that a little bit further, we’d say 1 over 9. A fractional exponent like 1 over n means to take the nth root. So for 16 to the 1/2 power, it would be the square root of 16, which equals 4. Finally, a fractional exponent like m over n means to take the end through first then raise it to the nth power. So here this will mean to take the square root of 4 but then raise that to the 3edpower. The square root of 4 would be 2 so 2 to the 3rd power will be 2 times 2 times 2, which equals 8.
The last topic we’re going to talk about is absolute value. The absolute value of a number is a number’s distance from 0 on a number line. A quick way to remember how to do absolute value is that it is always positive. The absolute value of a number A is written like so. We have two little sticks on the outside of your number. So if you’re looking at the absolute value of 4, here’s 4, here’s 0, that would be 4 spots away. So the absolute value of 4 is 4. If you’re looking at -4, once again, here’s 0, that will be one, two, three, four spots away. So the absolute value of -4 is also 4. Now the distance between two numbers on a number line is always positive. Take the absolute value of the difference. So if we want to find the difference between -5 and 6, we’d say first let’s subtract -5 minus 6, that’s the same thing as adding the opposite, so -5 plus -6 will be -11, but if we want the distance, so it’s the absolute value of that, which means the distance between -5 and 6 is 11.
Now let’s move on to fractions. When adding and subtracting fractions, you need to find a common denominator. Usually, you’re trying to find the least common denominator, or the LCD. That means it’s the least common multiple of the denominators. So looking at 2/3 and 4/5, we need to find the least common multiple of the denominators, which would be 3 and 5, and that would be 15. But we need to change our numerators. We can’t just leave them as 2 and 4, so what we’re going to say is our original denominator was 3, 3 times 5 gave us the 15, so 2 times 5 gives us a new numerator of 10. For 4/5, 5 times 3 gave us 15, so 4 times 3 gives us 12. And now, since we’ve the same denominator, we can add our numerators, our denominator stays the same and we want to make sure we change that into a mixed number if we have an improper fraction. So our answer will be 1 because 15 goes into 22 once and there will be 7/15 left over.
When multiplying fractions we want to cross reduce and divide out any common factors, then you’re going to multiply your numerators and multiply your denominators. So cross reducing means looking on the diagonal. Here 3 goes into both 3 and 9, so we’re going to divide that out. So 3 goes into 3 once, and 3 goes into 9 3 times. On this diagonal we can divide out of 5. Five goes into 5 once, 5 goes into 10 twice. So now we’re going to multiply 1 times 1, which gives us 1, and 2 times 3, which gives us 6, so our answer is 1/6. For division, you multiply the first fraction by the reciprocal of the second fraction. That really means you’re going to keep change flip. So we’re going to keep 70 as the same, our division changes to multiplication, we take the reciprocal of the second fraction, which means we flip our numerator and our denominator. So we’re going to say 20 over 13. So now that we’re multiplying, we can follow these rules and we can cross reduce. I see that 4 goes into both 8 and 20, so 4 goes into 8 2 times and 4 goes into 20 5 times. Nothing divides out between 7 and 13, so we multiply our numerator 7 times 5, which gives us 35, 2 times 13 is 26, and if we change that back to a mixed number, it will be 1 and 9/26.
If you have mixed numbers, you should change them to improper fractions before you start trying to solve the problem. We’re not going to totally finish this problem but if we’re going to start it off, we would say 2 and 1/5 well, 2 times 5 is 10 plus the 1 is 11, so 11/5, 5 and 3/7, 5 times 8 is 40 plus 3 is 43, so 43/8 and then you’ll continue on from there.
For decimals, when you’re adding and subtracting, you need to line up the decimal point and then you’re just going to add or subtract as usual. So here we’ll say 3.34 minus 1 point, notice the decimal points are lined up, 1.715 but there’s a missing space here, but if we’ve a terminating decimal, which is the case, we can always add a 0 to the end because that doesn’t change the decimal. And now we can subtract. We can’t do 0 minus 5 so we need to borrow, so the 4 becomes a 3 and then 0 becomes a 10, 10 minus 5 is 5, 3 minus 1 is 2, need to borrow again, so that becomes a 2 and that becomes 13, 13 minus 7 is 6, we just bring down a decimal point and 2 minus 1 is 1.
For multiplication we don’t want to line up the decimal points. We’re going to set up the numbers as if are doing a long multiplication problem, but line up the digits of your numbers along the right-hand side. So looking at 3.8 and 2.19 I see that this 1 has more digits, so I’m going to put that one on top. We don’t want to put 3.8 here because that’s going to make life a little bit harder for us, so we’re going to line up the digits along the right, the digits line up. Now we can multiply, 9 times 8 is 72, 8 times 1 is 8 plus 7 is 15, 8 times 2 is 16 plus 1 is 17, leave a space, 3 times 9 is 27, 3 times 1 is 3 plus 2 is 5, and 3 times 2 is 6. We add them together, we get 2, we get 12, we get 13, we get 8, and now we’re going to count how many decimal places were in our question. We had two decimal places here and one here so our answer is going to have three total decimal places so one, two, three, so 8.322. Finally, for division, if we’re dividing by a decimal, we want to slide the decimal over in our divisor until we get a whole number and then whoever many time as we slid it over we’re going to do this same thing in the dividend. So we’re going to do 1.43 divided by 1.1. However, we can’t divide by a decimal, so we slide it over once here, so we have 11, which means we’re going to slide it over here, so 11 goes in the 14 once with the remainder of 3, we bring down our 3, 11 goes in the 33 3 times, with no remainder, so our answer is 1.3.
So let’s work with squares and square roots. Squaring a number means you’re multiplying a number by itself, it means you have the power with an exponent of 2. So 5 squared would be 5 times 5, which equals 25. Negative 8 quantity squared would be -8 times -8. We learned that a negative times a negative is a positive, so it’ll be a positive 64. This question looks like our last one, however, there’s no parentheses, so what this means is we’re doing 8 squared but then on front of it we’re going to stick a little negative sign. So this means -8 times 8, which would mean that this is 64 with a negative out front, and then 2/3 squared would be 2/3 times 2/3, multiply our numerators, multiply our denominators, and we get 4/9. Finding the square root of a number means finding a number that what number multiplied times itself equals the given number. So the square root of 81 means we need to find a number that when you multiply by itself gives you 81, and the answer to that is 9. The square root of 121 means what number times itself equals 121. The answer to that is 11. Now a lot of times the number under the square root is not a perfect square, 81 and 121 are examples of perfect squares because when you square root it you get a nice whole number.
If we start from the beginning, our perfect squares would be 1 squared, which is 1, 2 squared, which is 4, 3 squared, which is 9, 4 squared, which is 16, 5 squared, which is 25, and so on. But most of the time, like I said, these numbers are not perfect squares, so we can leave these answers in radical form, so what we’re going to do is see if we can find a perfect square that goes into our numbers. Well I know for 32 that that equals 4 times 8. So we can say that root 32 is really root 4 times root 8, but really root 8 could be broken out more, so that’ll be root 4 and then root 8 will be root 4 times root 2. And now we have the square root of 4, that’s 2. Again, the square root of 4 is 2, and then we can’t really do anything without root 2, so we leave it the same. So our final answer will be 4, this 2 times 2 is 4, 4 root 2.
Root 80, let’s see, root 80 would be root 4 times root 20, but root 20 would be root 4 times root 5 so this will be 2, this will 2, leave that as root 5, and our answer is 4 root 5.
Now let’s work with cubes and cube roots. Cubing a number means you have a power with an exponent of 3. So 5 cubed will be 5 times 5 times 5, which is 125. Negative 2 quantity cubed will be -2 times -2 times -2, these here give you a positive 4, so a positive 4 times a -2 is a -8. Negative 2 cubed would be -2 times 2 times 2, this here gives you 8, so it’ll be -8. And then 2/3 cubed will be 2/3 times 2/3 times 2/3, 2 times 2 times 2 gives us 8, 3 times 3 times 3 gives us 27, so its 8/27.
So taking the cube root of a number means finding a number that when you multiply it by itself two other times, you get whatever number’s underneath it. So here we want to find a number that when you multiply it by itself again and again one more time, gives you 8. Well, 2 times 2 will be 4 and times 2 again will be 8. So the cube root of 8 is 2. The cube root of 1000, if we took 10 and multiply that by 10 will be 100, and 100 times 10 again will be 1000. So the answer for the cube root of 1000 is 10.
Now there’re times when numbers are undefined. Undefined means when you have m over 0 and m doesn’t equal 0, that gives you something else. So if I have 8 over 0, it’s undefined. This is because the fraction bar really means division and we can’t take 8 things and divide it up into 0 groups, that just doesn’t make sense. So whenever you’ve a fraction that has a denominator of 0, it’s undefined. Now certain numbers are undefined in certain sets. So square roots of negative numbers are undefined in the set of real numbers, you can’t have a square root of a negative number because a negative times a negative will give you a positive, not a negative number. There are imaginary numbers. That will make this problem possible, however, in a set of real numbers square root of -81 is undefined.
Finally, let’s work a little bit with scientific notation. Scientific notation is a way of writing really big numbers or really tiny decimals in a shorthand way, so you don’t have to write out all the zeros inside the numbers. So if we want change from standard form to scientific notation, we want to place a decimal somewhere in our number so that we create a number between 1 and 10. So where we would put this decimal here to create a number between 1 and 10 will be right there between the 9 and the 2, because 9 is in between 1 and 10. So we’re going to rewrite this, we’re going to say 9.28, we chop off all those extra zeros, you’re always going to have times 10 to some power.
Now it’s important to remember this here, a positive exponent is going to give you a big number. If you put a negative exponent it means you’re talking about a decimal. So here was a big number so we’re going to have a positive exponent. And to determine where that exponent is going to be, we’re going to start with our original decimal and see how many spots we slid to get to our new decimal. Well, you’ve to remember that on a whole number there’s an invisible decimal at the end of it. So we slid it once, twice, three times, four, five, six. So it’ll be 9.28 times 10 to the 6th power. For this once, since it’s a decimal we know it’s going to have a negative exponent. So we need to find a number between 1 and 10. That would happen by placing the decimal between the 6 and the 2 and then we’re going to count how many we moved. So one, two, three, four, five. So 6.2 times 10 to the -5th.
To go back from scientific notation to standard form, what we’re going to say let’s take that 2.78 and write over here to give ourselves a little bit more room. And since it’s 10 to the 7th power we want to move it 7 places. Now since it’s a positive 7, it means we’re creating a big number, so we want to slide that decimal point to the right. So we’re going to start here then move it 7 times, so one, two, three, four, five, six, seven, and with all those blank spots, we’re going to fill in zeros. So our final answer will be 27,800,000. Let’s rewrite 4.15. We need to move it five decimal places and since it’s a negative number we’re creating a decimal, which means we move it to the left. So we’re moving it once, twice, three, four, five. We fill in those decimals, notice it’s not how many zeros we add to our number, it’s how many times we slide that decimal place, so our answer is 0.0000415.
Finally, let’s multiply two numbers in scientific notation. We have 2.7 times 10 to the 7th and 3.1 times 10 to the 5th. Now multiplications commutative and associative means we can change up the groups and we can do whatever order we want. So let’s group 2.7 and 3.1 together, and group 10 to the 7th times 10 to the 5th together, 2.7 times 3.1, 3,7 times 2 and 3 times 7 is 21, 3 times 2 is 6 plus 2 is 8, one decimal place here, one decimal place here, which means two decimal places is our answer, so this gives us 8.37 and we learned that when we multiply the same base we add our exponents together, so it’ll be 10 to the 12th power.
The last thing we’re going to do is calculate weighted average. According to a teacher’s syllabus, 60% of Anna’s marking period average is based on her tests and quizzes, 30% is based on homework, and 10% is based on classwork. She has a testing quiz average of 85, a homework average of 95, and a classwork average of 90. What is Anna’s marking period average rounded to the nearest whole number? So her tests and quizzes were 60%, and that was a grade of 85, 30% was homework and her homework grade was a 95, and then 10% is classwork and her classwork was a 90. So what we’re going to do is multiply the percent by the respective grade. However, when we multiply by percents, we have to remember to change the percent to a decimal. So what we’re really doing here is .6 times 85, .3 times 95, and .1 times 90.
And then we’re going to add those up together, so 85 times .6, 5 times 6 is 30, 8 times 6 is 48 plus 3 is 51, one decimal place, so the first part is 51; 95 times .3, that gives you 15, that gives you 28 with one decimal, so 28.5; and then .1 times 90 is just going to be 9. You can fill in those missing decimals. So we add, we get 5, we add 18, we get an 8, and round it to the nearest whole number. Anna’s final marking period grade is going to be an 89.
So the four main measures of central tendency are mean, median, mode, and range. When you hear the word average, that’s talking about the mean. You add up the numbers and divide by however many numbers you had. The median is the middle number, you need to order your numbers and then find the one in the middle. Now if there’re two numbers in the middle, you’re just going to find the mean of the two of them. So you add those two numbers up and divide it by 2 and that’s your median. The mode is the number that occurs the most often, and the range is the biggest number minus the smallest number. So we’re going to find the mean, median, mode, and range of this set of data.
So for mean we’re going to add up the numbers, 4 plus 11 gives you 15, plus 2 is 17, plus 5 is 22, plus 6 is 28, plus 4 is 32, plus 3 is 35. So the sum was 35 and we’re going to divide by however many numbers we had, which was 7, which means the mean is 5. For median we’re going to put our numbers in order. So our smallest number is 2, then 3, then two 4s, then a 5, a 6, and 11. A lot of people like to cross numbers off. So you start on one side, you cross off a number on the other one. You keep going crossing of the same amount on each side and then you get your number in the middle, which is your median. The mode is the number that occurs the most often, well, the number 4 occurs twice while everything else only appears once, which means that 4 is our mode. And the range is the biggest number minus the smallest number, so that will be 11 minus 2, which gives you 9.
Now often you’ll be given a set of data, you’ll be told what the average is and then you have to find a missing part of the data. So for example Dan has received test grades of 90, 80, 65, and 95. What grade does he have to get on his fifth test to ensure his test average is an 85? So you need to figure out the goal amount of points that Dan needs to get. Well if the average is 85, then we can say let’s take the 85 and multiply it by the 5 tests. Five times 5 is 25, 8 times 5 is 40, plus 2 is 42. So altogether his test grades need to add up to 425. Well, let’s figure out what he has so far. We’re going to add up the ones that we’re given, it gives us 10 that’s 18, plus 6 is 24, plus 9 is 33. So, so far he has 330 points. He needs 425, he has 330, we’re going to subtract to figure out how many he needs on that last test. We get 5, we borrow, so Dan needs to get a 95 on his fifth test to make sure his test average is an 85.
Now we’re going to find the average number of siblings based on the frequency counts provided in the table. Now it’s often good to take the data and write it out so we can see it a little bit better. So 0 occurs once, 1 occurs four times, 2 occurs twice, 3 occurs twice, and 4 occurs once. The average means we’re going to add them up and divide by 10 because there’re 10 numbers, right here is 4, plus another 4 is 8, plus 6 is 14, plus 4 is 18. So the sum is 18. We’re going to divide that by 10, and the average is 1.8.
Let’s deal with a little bit of probability now. Probability is the number of favorable results over the total possible results. And probability is always going to range between 0 and 1. Zero means it’s definitely not going to happen. A probability of 1 means it’s definitely going to happen. But most of the time our probabilities are going to fall somewhere in between the two. You can write probability as a fraction, a decimal, or as a percent, but you have to make sure it’s between 0 and 1. So if we roll a six-sided die once, what is the probability of rolling an even number? If we have a six sided die that means the total possible results are six, because in there are 1, a 2, a 3, a 4, a 5 or 6.
Well, even numbers would be 2, 4, and 6. So those would be the favorable results so we can have 3 possible good things over 6 total results which means our probability is 1/2. Now a lot of times you’ll have to find probabilities for several things happening at once. That’s what is called a compound event. It’s a combination of two or more simple events and can be found by multiplying the two probabilities together. Now two things can happen. You’re going to have independent events, which is events where one outcome does not affect the other outcome. On the other hand, dependent events means that a result of the first outcome affects the probability of the second outcome. So we have both cases demonstrated here. If you roll a six sided die and flip a coin, what is the probability of rolling a number bigger than 4 and having the coin land on heads? Now if you’re rolling a six sided die and flipping a coin those two things are totally separate. Flipping a coin is not going to affect rolling a die and rolling a die is not going affect flipping a coin. So these are independent events, so we can just figure out each of the probabilities and then multiply them together to get our answer.
So for the die, we want a number bigger than 4. Once again for a die, we have six total possible results. If we want a number bigger than 4 our options would be 5 or 6, which means there’re two favorable results. For a coin if we wanted to land on heads, there’re only 2 possible options: heads or tails, and heads would only be one favorable outcome. So we have 2/6 times 1/2, we’re going to multiply those together, we can either simplify and now or simplify later, but let’s multiply, we get 2 over 12 and then we can simplify that to 1 over 6, and that is the probability of rolling a number bigger than 4 and having the coin land on heads.
Well, that was an independent event, let’s look at some dependent events. There are 4 yellow marbles and 6 blue marbles in a bag. What is the probability of choosing 2 blue marbles without replacement? When you see without replacement, that usually clues you in to the fact that it’s dependent and you need to be careful, because if you take out that first blue marble, there’s not going to be the same number of the marbles in that bag, so the probability of taking that second marble is going to be affected by taking out the first marble. So we’re going to say let’s assume that we reach our hand into this bag and pull out a blue marble. Well that probability 4 plus 6, there are 10 total marbles at first, and there’s 6 blue ones. So our first probability of grabbing a blue marble is 6 over 10.
So now we need to assume that we have taken that marble out and we’ve a blue marble, which means that the number of marbles in our bag has gone down from 10 to 9, and if we took out a blue marble we don’t have 6 anymore, we have 5 instead. So let’s cross reduce in this case. We’ll simplify before we multiply, 5 goes in the 5 once, 5 goes into 10 twice, 3 goes into 6 twice, and 3 goes into 9 3 times. Actually I can simplify here more, they reduce, so really just 1 times 1 is 1, and 1 times 3 is 3, and that is our dependent probability.
The fundamental accounting principle says that if there are m ways for one activity to occur and n ways for a second activity to occur then multiplying m times n will give you the total possible outcomes of both occurring. So for example, a restaurant offers a three course meal where you can choose one appetizer, one entrée, and one dessert. There are 5 appetizers to choose from, 8 entrees, and 3 desserts. How many possible three-course meals can you choose from? In this case all we have to do is take the number that we have in each category and multiply everything together. So we have 5 appetizers, 8 entrees, and 3 desserts, so we multiply 5 times 8 times 3, 5 times 8 is 40, and 40 times 3 is 120. So there’re 120 possible meals to choose from.
Some other counting techniques are permutations and combinations. These are both ways of ordering sets. The important question you need to ask yourself is does the order of the set matter? For permutations, the order matters. Some examples are races, finishing first, second, third, choosing officers such as the president, vice president, secretary, and treasurer, trying to calculate pin numbers, and arranging books on a shelf. Two things can happen with permutation: we can either order everything in the set, or we can just choose to take a couple of things from the set and order those.
In our first example, it says in a race of 5 runners, how many ways can the runners finish? So in this case we’re ordering all 5 runners and to do that, we’re going to take the number we have, which is 5, and we say 5 factorial. Now what a factorial means is you’re going to take the number that you’re starting with and you’re going to keep multiplying by every number that’s smaller than it until you get to 1. So 5 factorial means 5 times 4 times 3 times 2 times 1. Five times 4 is 20, 20 times 3 is 60 and 60 times 2 is 120. So they can place in 120 different ways.
Now, we’re going to have a similar question but in this case we only care about first, second, and third place. It says in a race with 5 runners, how many ways can the runners finish first, second, or third? In this case we’re not ordering all 5 of the runners. Just, couple of them. In your calculator there’s often a button that will do this for you. In that case, we would say we’re taking our 5 runners, there should be some function for permutations, it’s usually NPR, and if we’re choosing 3 people we would say 3 and the calculator will do it for us. However, right now we don’t have a calculator so we’re going to do it the old-fashion way.
And that’s using this formula here. We’re going to take n factorial, which is how many runners we have, so 5 runners, and then this n minus k is the number of runners minus the number of people we’re actually taking in our set. So it will look like 5 factorial over 5 minus 3 factorial, which is really 5 factorial over 2 factorial and we’re not actually going to multiply anything, I’m just going to write it out so 5 times 4 times 3 times 2 times 1, and then over here in the bottom you have 2 times 1, and in these cases of permutations and you’ll find that the same thing happens in combination, we can often cancel out some common factors in our numerator and our denominator. So over here we have 2s and 1, so we can cross off, which means we are really just left with 5 times 4 times 3. Five times 4 is 20 and 20 times 3 is 60. So the runners can finish first, second, or third in 60 different ways.
Now for combination the order does not matter. For example, if you’re choosing a committee, which is just a group to represent people where no person has a different role than everyone else, that’s a combination, because it doesn’t matter in what order you choose those people. You might also see it picking pairs or groups, that’s a combination. So let’s look at this example. How many ways can 2 names be chosen from 10 in a raffle if only 1 entry per person is allowed? So in this case we’re starting with 10 but we’re choosing 2 names. So once again, if we had a calculator we would say we’re taking 10 people, the button you often press is NCR, C for combination this time, and 2 people at a time. But let’s use our formula. It’s very similar to our permutation formula except we’re adding an extra k factorial at the end. So let’s say 10 factorial over 10 minus 2 factorial and our k is 2 factorial. That’ll be 10 factorial, 10 minus 2 is 8, and now let’s write it down, we have 10 times 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 and then 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1, and then another 2 times 1. And well, it looks like a lot, we see we have an 8 and an 8, 7 and a 7, and a lot of our numbers will divide out. So really we’re left with 90, 10 times 9, over 2 times 1 which is 2, and 90 over 2 is 45.
This is a GED review video for expressions and polynomials. First we’re going to deal with linear expressions. And these are expressions that have one variable and that variable needs to be raised to the power of 1. Now first we need to deal with some definitions. Terms are the parts of an expression that are added together. Like terms are terms that have the same variable raised to the same exponent. For example, if I have 2x and 3x, those are like terms because they both have a variable of x and both xs are raised to the first power. Finally, a constant is a number with no variable attached. For example, 4. In an expression, two constants are also considered like terms. So first we’re going to add two expressions together. So I have 4x plus 7 plus 8x minus 4. Now the only thing we can do here is combine our like terms. And once again, like terms are terms that have the same variable raised to the same power. So here I have 4x and 8x and also have a positive 7 and while this says minus 4, we should know that minus and negative are really the same thing because we could change this to plus a -4 if we chose to do so. So we’ve a positive 7 and a -4. So 4x plus 8x, we add the coefficients together, 4x plus 8 x gives us 12x, and 7 plus a -4 is a positive 3. So our answer is 12x plus 3. Now if we’re subtracting, we have 4x plus 7 minus 8x minus 4. Now when you’re subtracting expressions, you have to remember that this minus sign actually means we’re subtracting both things in our parentheses.
So it might be helpful to rewrite the problem. We have 4x plus 7, but now we’re subtracting 8x, and here we’re subtracting a -4, which really means we are adding 4. So once again, my like terms are 4x and in this case -8x, and then 7 and positive 4. 4x plus -8x gives me a -4x, and 7 plus 4 is a positive 11. Here we have an expression on the inside multiplied by a whole number on the outside. This is what’s called the distributive property. We need to take this number on the outside and multiply it by both things on the inside. So we’re going to say 3 times 4x. We always bring down whatever sign is there, and then 3 times 7. Well, 3 times 4x is 12x, and 3 times 7 is 21. Once again, not like terms. So that is our final answer.
Now I just want to back up one second because there’s actually another way of writing each of these answers. It’s usually common to write your answers with your variable first and then your constant after. Or in certain cases where we have multiple terms going on, we usually start with the biggest exponent and work our way down. However, it’s also possible to switch the order of our terms. So for our first one, instead of saying 12x plus 3, we could say 3 plus 12x. For our second one, we need to make sure that the sign in front of it stays with it when it moves. So if you want to move this 4x, we need to keep that minus sign in front of it, and this positive 11 is going to go out front and since it’s upfront we don’t need to put the positive sign in front of it. And finally for this one, there’s no negatives to worry about, so we can also say if we chose 21 plus 12x.
We now have an example of factoring and this example that we have is kind of the most basic type of factoring we could do. It’s pretty much the reverse distributive property. Here we had a number on the outside and we multiplied it to both things on the inside to get rid of our parentheses. Now we want to see if there’s a number that goes into both of our terms, we’re going to pull it out and kind of create some parentheses here. So I have 8x minus 4, and I can say that 4 can go into both 8x and 4. So I’m going to pull out the 4, which means in my parentheses I just say 4 times what gives me 8x? And the answer to that will be 2x. Okay, bring down our minus, 4 times what is 4, and that answer is 1. So 8x minus 4 factored will be 4 times in parentheses 2x minus 1.
This is the GED review video for expressions and polynomials. You often need to evaluate expressions by substituting an integer in for an unknown quantity. So for example we’re going to evaluate 4x plus 7 for x equals 3. I would recommend writing in your numbers first, especially when your expressions are a lot bigger, so you don’t make any careless mistakes. So we’re going to say 4, this means multiplication when they’re squished together, so 4 times 3 plus 7 following order of operations means the multiplication comes first, so 12 plus 7, that gives you 19. This one here is a little bit trickier. We have a -x, so we write down our negative from here but our x happens to be -8, and then minus 7. Now that -8, the two negatives really make it a positive, so positive 8 minus 7 and our answer is 1.
This is the GED review video for expressions and polynomials. Now let’s evaluate a polynomial. Same idea as before. We’re first going to rewrite our expression but stick in the number that they give us. So -3 times 2 squared plus 7 times 2 minus 3. PEMDAS says exponents first, so we’re doing that, so -3 times 4, we bring everything else down. Now we can do our multiplication, -3 times 4 is -12, 7 times 2 is 14, and then bring down your minus 3, -12 plus 14 is 2, and 2 minus 3 is -1.
This is the GED review video for expressions and polynomials. Here we’re moving onto polynomials, which just means that we keep getting some more terms in our expression. If there’re two terms, it’s what’s called a binomial, three terms will be a trinomial, and anything more than that is just generally a polynomial. So here we’re going to add two trinomials together. Our like terms are 4x cubed and -6x cubed, we have 2x squared but no 2x squared over here. We have a -7 and positive 3, and then a -4x. So putting them together, 4x cubed plus a -6x cubed is a -2x cubed. There’s nothing to pair up with our 2x squared, so that stays the same. There’s nothing to pair up with our -4x, so that also stays the same, and then -7 plus 3 is a -4.
Okay, here we are subtracting so let’s rewrite it. We have -2x cubed plus 5x squared minus 7, but now we’re reversing the sign of everything in here, minus a negative will be plus an 8x cubed, minus a -2x will be plus a 4x, and then minus a positive 3, it’s the same thing as just subtracting 3. We have our -2x cubed and our 8x cubed, we have a 5x squared and nothing to match up with that. We have a -7 and a -3 and then a 4x. So -2x cubed plus 8x cubed is 6x cubed. Once again nothing goes with our x squared, nothing to match up with our plane x, and then -7 minus 3 is -10.
Now we’re going to multiply two polynomials. We need to follow what’s usually called foil, and that’s the answer for first, outer, inner, last. So we need to make sure we hit every pair of numbers that we have. And that’s where this comes into play. So the f stands for first and then we’re going to multiply the first numbers in each of our parentheses. So we have 4x times 8x, that’s going to give us 32x squared. Now we do the outer, so that this number and that number there on the outside edges, so 4x times a -4 is -16x. Now we do the inners. Start here and move on to there, 7 times 8x is 56x, and then we do this times the last number, 7 times -4 is -28. And really what we have to do now is combine our two middle terms, so we just bring down our 32x squared, -16x plus 56x is positive 40x, and then minus 28.
This is the GED review video for expressions and polynomials. Factoring. We started one before, example 1 is a little bit similar to that, so we’re going to try and do the reverse distributive property. So we’re going to see what goes into both of our terms. We can pull out a 2 and we can also pull out an x. So what times 2x will give us 6x squared? Well 2 times 3 gives us a 6, and x times x will give us the x squared, minus 2 times 2 will give us the 4, and we have x and x, so we don’t need any other xs. So this is our factored answer.
Example 2 is a trinomial, so it’s a little bit different. There’s nothing that goes into each of our terms but what we can do is kind of like a reverse foil. We’re going to create two sets of parentheses. This one is not bad because in order to create x squared, it’s just going to be x and x, because x times x gives us the x squared. But now we need to find two numbers that are going to multiply to -8 but then also add to -2. So our options for our -8 will be 1 and 8 but none of those will work adding to -2, and we could also have 2 and 4, and let’s say if we make that 4 negative, 2 times -4 will be -8, and 2 plus -4 will give us -2. So we’re going to take these numbers and fill them in over here. So we’re going to have an x plus 2, because a positive 2, and then an x minus 4, and that is our factored answer. The good thing about factoring is you could always go back and check to see if you did it correctly. If we now foil this, we should get back our original question. Sox times x is x squared, x times -4 is -4x, 2 times x is 2x and 2 times -4 is -8, so we get x squared minus 2x minus 8.
Example 3 is very similar to example 2 but the problem is we don’t just have x squared upfront, so you have to play around with it a little bit, it’s kind of guess and check, but it’s quite possible that we might have x and 6x in our parentheses, and then we need to figure out two numbers that multiply to -15 and then add to -1. But in this case we need to keep in mind that the numbers that we choose may also be multiplied by other numbers here, so it’s not exactly like this. We have to plug the numbers in and play with them. So maybe we could try -3 and positive 5 because those multiply to -15. But what happens is we’ll get a positive 5x here and a -18x here, which would be a -13x. So that doesn’t work. Maybe we could do 1 and 15, we’ll say this one is negative, this one is positive. In this case we’d have a -15x on the outside and a positive 6x on the inside, so that would add to -9x, which isn’t -1x. So I guess that you have to play around with it. We can also do 2x and 3x. Let’s try 3 and -5 and see if that works, 2x times 3x is 6x squared, 2x times -5 is -10x, 3 times 3x is 9x, and 3 times 5 is -15. And then these two together will give you your -x. So this is the correct factored answer.
In this case we have a polynomial with four terms. Nothing goes into each of them. This method won’t work because it only works for trinomials, but in this case we can kind of look at our pairs of terms and see if we can pull anything out. So I’m going to treat these as a set and I’m going to treat these as a set. Looking at 3x cubed minus 2x squared, I can actually pull out an x squared. That will mean that I’d have an extra 3 and an extra x and then over here I’d have an extra 2. We’re now looking at my blue set, I have 12x minus 8, I could pull out a 4, which means I’d have 3x left and I’d have a 2, and now from each of those sets we have actually pulled out the same binomial. So what we’re allowed to do is we can say well, one of our pairs of factored terms is 3x minus 2, and then we’re going to combine our extra stuff together. So we’re going to multiply that by x squared plus 4. And finally there’re certain special cases, this is one of them, when you have binomial but in the side binomial we have perfect squares. What we can do is we can say let’s take the square root of each of them, so the square root of x squared will be x, the square root of 16 will be 4, and one of them is going to be plus and one of them is going to be minus, because what happens there is we have a positive 4x and a -4x, and our middle plain x terms disappear.
Okay, now we have a polynomial in our numerator and our denominator, so what we can do is we can factor both of our polynomials and see if we can simplify the expression at all. So this one here on top we have x and x, we need to find two numbers that multiply to -8 and add to -2, so that would be -4 and 2. On the bottom we can pull out a 2, which will be x minus 4 left. And then since we’ve an x minus 4 in the numerator and an x minus 4 in the denominator we can cross them off, and we are left with x plus 2 over 2.
This is the GED review video for expressions and polynomials. Now we’re going to add and multiply rational expressions. So here we have 3x plus 7 over x plus 5, plus 2x over x minus 3. It’s just like if we’re adding fractions with just numbers. We need to find a common denominator. And in this case, our common denominator is going to be x plus 5 times x minus 3. All you need to do is multiply your two denominators together. However, whatever we do to the denominator, we need to do to the numerator. So that means in this case for our first term, if we had x plus 5, we need to multiply it by x minus 3, which means we multiply our numerator by x minus 3 also. So we foil, we get 3x squared minus 9x plus 7x minus 21. And that will be 3x squared, these two together give you minus 2x minus 21.
So that’s our first fraction. Our second fraction we’ll need to multiply it by x plus 5, which means you need to multiply the numerator by x plus 5. We get 2x squared and 2x times 5 is 10x, so our second fraction gives us 2x squared plus 10x over x plus 5 times x minus 3. So we’re going to keep going and we’re going to combine our like terms together, so 3x squared plus 2x squared is 5x squared, -2x plus 10x is 8x, and then minus 21, and then we have x times x. We’re just going to now foil this out so we know what it is, x times x is x squared. We’d have a -3x and a positive 5x, so that’s a positive 2x, and then a 5 times 3 is a -15, and that is our answer.
For multiplying, we can first factor our polynomials and see if we can divide out any common factors. So we’re going to have 4x, nothing to factor here, just 2x plus 3, however, on the top over here we can divide out a 2 and that’ll be 2x plus 3 over x. So our 2x plus 3 will cancel, our xs will actually also cancel, and all we’re left with is 4 times 2, which is 8.
This is the GED review video for expressions and polynomials. Last example, all we’re going to do here is once again substitute our y in and evaluate the expression. So we’ve 4 squared minus 4 over 3 times 4. This gives you 16, on the bottom we have 12, 16 minus 4 is 12, and 12 divided by 12 is 1.
This is the GED review video for linear equations. There are certain steps we should follow when solving equations. Not all equations will need to have all of these steps but it’s good to know what order you should do things in. The first thing is if you see parentheses, you need to distribute. You should then combine any like terms that are on the same side of the equal sign. And notice there’s no inverse operations. Inverse operations is just the opposite operation. When we’re trying to move things from side to side, that’s when we’re going to do the opposite, but if we’re just doing on the same side there’s no inverse operations needed. Next thing is going to be to get all your variables to one side of the equation and all constants to the other side, and that’s when you’re inverse operations will be needed. And finally, we’re going to multiply or divide to get the variable by itself.
Now the first example is very basic, x minus 8 equals 10. To get rid of minus 8 we must add 8. Whatever we do to one side we need to do to the other side, so we get x equals 18. Over here -7 minus x equals -5, we need to get rid of the -7. To get rid of -7 we must add 7 and the reason we’re doing the inverse because -7 plus 7 equals 0. And if you have 0 with your x, then it’s just going to be your x. In this case though, we actually have -x, so we need to bring -x down, and we have -5 plus 7 which is 2, and the problem is we don’t want -x, we want positive x, and we’re allowed to switch that sign as long as on the other side we also switch the sin.
Our next example is a two-step equation. Now we have variables on this side, which means we should get all their constants to this side. To get rid of minus 8 we’re going to add 8, so we get 3x, -26 plus 8 is a -18. This right here is a multiplication so inverse will be to divide, and by dividing we get 1x, which is just x, and x equals -6. Here if we’re looking at the left-hand side we’ve a lot going on. So we need to combine our 4x and our -10x. And notice since they’re on the same side of the equal sign, we’re just doing what the signs tell us to do, there is no inverse operations yet. So 4x minus 10x is -6x plus 5 and that equals -23.
Two step equation, let’s get our constants to the right side. To get rid of plus 5 we subtract 5, so we have -6x, equals -28. This actually isn’t going to divide into -28 evenly, however, that’s not always going to be the case, so let’s practice. We’re still going to get rid of our multiplication by dividing, and we can either divide this out and leave it out as a decimal, or we can leave it as a mixed number. Our answer is going to be positive because a negative divided by negative is a positive, so we don’t have to worry about those, 6 goes into 28 4 times with 4 a leftover, so x equals 4 and 2/3.
And in our last example on the page, we need to simplify both sides of our equation. Over here we need to distribute, and on the right hand side we need to combine like terms. So 3 times x is 3x, 3 times 7 is 21. On the other side, 9x minus 2x is 7x, and then we have our minus 3. Now we need to get all of our variables to one side and our all constants to the other side using inverse operations. We actually have two options here: we can either get these 3xs over here or these 7xs over there. We could say let’s get rid of 7x by subtracting it, which isn’t bad, but in this case it’s going to make my coefficient negative. So what I try to do is I try and rearrange it so that my x stays positive. So instead, I’m going to say let’s get this 3x over there by subtracting 3x, which means we have 21, 7x minus 3x is 4x, all my xs are on the right side, we still have minus 3. Since my xs are on the right side, I need my constants on the left. So to get rid of minus 3 we’re going to add 3, so 24 equal to 4x and the final step is to divide by 4, and x equals 6.
This is the GED review video for linear equations. We can sometimes use equations to help us solve word problems. The total cost of repairing a car is the sum of the amount paid for parts and the amount paid for labor. You paid $82 for parts and $40 for each hour of labor, the total cost to repair the car was $242. How many hours did it take to repair the car? So first we should identify the variable, and the variable is going to be what the question is asking you to find. In this case it’s asking us for hours, so we can say that h equals a number of hours. Now the numbers that need to somehow go into our problem are 82, 40, and 242. It’s often good to find a total if there’s one. In this case the total is 242, so we know we’re going to be adding something together that’s going to equal 242. Now in our equations there’s usually going to be a number multiplied by the variable. So you can look at the numbers you have and see which one doesn’t make sense to multiply by our variable. Would it make sense to multiply the cost for parts times the number of hours? No, so that 82 can just stay by itself. Would it makes sense to multiply the cost for each hour by the number of hours? That it would, so we can say 40 h, and that’s our equation. And now to solve, we can say minus 82 from both sides, 40 h equals 160. If we divide by 40 we get 4 hours.
You spent $40 on clothes and bought 4 DVD movies. Your friend spends nothing on clothes and buys 8 DVD movies. You both spend the same amount of money. That’s important. All the DVDs cost the same amount. How much does each DVD cost? So our question asked for the cost of the DVD, so we’re going to say c equals cost of DVD. Remember you can use whatever variable you’d like. So in this scenario we have one friend buying something, and the other friend buying something. Those two friends spent an equal amount of money. I have to deal with 40, 4, and 8. Doesn’t make sense to multiply the cost of close times the cost of a DVD? No, they have no relationship. Would it make sense to multiply the cost of a DVD times the number of DVDs you bought? That it would. So plus 4c. Your friend spends nothing on clothes and buys 8 DVD movies, so that will be 8c, and that’s our equation. Since we only have a number over here, we should probably get our variables to this side, so we can subtract 4c, we get 40 equals 4c. The final step is to divide by 4, and each DVD costs $10.
This is the GED review video for linear equations. The last thing we’re going to talk about is systems of linear equations. That’s when you have two equations with two variables and you need to find the x and the y that works in both of them. And there are several ways that we can do this. The first way is a linear combination. So here we have -3x plus y equals -3, and x plus 2 y equals 8. What we’re going to do is we’re going to take one of the equations and multiply it by some number so that if we add the 2 equations together either our xs will disappear or our ys will disappear. Here I have a -3x, so if I wanted the xs to disappear, I would need this to be a positive 3x, because -3 plus a positive 3 is 0. So I’m going to take my second equation, I’m going to multiply everything by 3.
Now my other option would’ve been to say well if I have 2y here, I can make this so that I have a -2y on top, that would also work. But we’re going to try get rid of our xs. So our first equation is going to stay the same, 3 times x is 3x, 3 times 2y is 6y, and 3 times 8 is 24. So now if I add these 2 together, -3x plus 3x is 0, y plus 6 y is 7 y, and -3 plus 24 is 21. And if I divide by 7, I get a y equal to 3. And now all you need to do is choose one of your equations, it doesn’t matter which. We can say x plus 2y equals 8. We’ll use the second one. We’re going to sub the y that we just got into that equation, so x plus 2 times 3 equals 8, which is really x plus 6 equals 8. One step equation, we subtract 6 from both sides, and x equals 2. So x equals 2 and y equals 3 is the solution to this system of equations. You may also see it written as an ordered pair, where your x is the x coordinate and the y is the y coordinate.
Here we have the same system of linear equations, but we’re going to figure out our solution by using substitution. Now substitution means you’re going to take whatever your equations and get one of the variables by itself on its own. And then you’re going to take that and plug it to the other equation. It doesn’t matter which one you do. Let’s use the bottom for now. So if we want x by itself, that means that 2y has to get over here, so we would subtract 2y. We get x by itself. And now we can’t combine these because they aren’t like terms, so it’s just -2y plus 8. So now we’re going to take our other equation, and we’re going to take what we just got and stick it in for the x. So -3 times -2y plus 8 plus y equals -3. This gives you 6y minus 24 plus y equals -3. Combine our ys together, so 7y minus 24 equals -3. We can add 24 to both sides, 7y equals 21. We can divide by 7, and y equals 3. We then need to plug this back in so we can find our x. So 2 times 3, x plus 6 equals 8, subtract 6, and x equals 2.
Finally, we can do this by graphing. Now I’m not going to go into whole lot of detail about graphing here so if you need to, please watch the separate video on graphing linear equations. So to graph these I first need to put them in slope intercept form. So my first equation, to get y by itself, I add 3x to the other side. So y equals 3x minus 3, that’s in slope intercept form. So this tells me the y intercept, so the y intercept is at -3. The number in front of the x tells me that slope, so I’m going to start at that y intercept and use a slope of 3, which means 3 up, 1 to the right, 3 up 1 to the right. And then we’re going to connect them like so. For my other one I need to subtract x from both sides, 2y equals -x plus 8 and then dividing by 2 means that we’ll have -1/2x plus 4. Intercept at positive 4, our slope is -1/2. So I’m starting here, that means I’m going to go 1 down, 2 to the right, 1 down 2, to the right, and connect my points. And if we look at where they intersect, the point they intersect at is (2, 3), which means that is the solution to this system of equations.