so i've decided to create this algebra review video because most of you are going back to school in about a week or two so let's review some basic concepts that you'll need in a typical algebra course first you need to understand how to add subtract multiply and divide fractions so let's briefly review that concept let's say if we want to add three over four plus two over five how can we do so now whenever you're adding or subtracting fractions you need to have the same denominator which we don't have for the first fraction the denominator is four for the second one we have a five in the bottom so we need to adjust these two fractions in order to get a common denominator we need to multiply the first fraction by five over five whatever you do to the top you must do to the bottom so if you're going to multiply the top number by five you have to multiply the bottom number by five now the second fraction i want to multiply the top and the bottom number by four because five times four will give us 20 and that will lead to the same denominator so now we could add the two fractions 15 plus 8 is 23 so the answer is 23 over 20. now here's another example 5 over 6 minus 4 over 7. feel free to try this problem if you want to go ahead and pause the video now if you're not sure what to multiply by multiply by the other denominator so five over six you can multiply the top and bottom by seven and four of a seven multiply the top and the bottom by six this will give you a common denominator of forty two now seven times five is thirty five four times six is 24 and 7 times 6 is 42 and now we can subtract the two numbers on top so 35 minus 24 that's 11. so the answer is 11 divided by 42. now what about multiplying two fractions what's seven over five times four over three when multiplying two fractions you don't need to have the same denominator in the case of when you're adding or subtracting fractions all you need to do is just multiply across seven times four is twenty-eight so you got to multiply the two numbers on top and then after that multiply the two numbers on the bottom five times three is 15. so the answer is 28 divided by 15. so here's another example 3 over five times six over four three times six is eighteen five times four is twenty now sometimes you may need to reduce the fraction which is what we need to do in this example eighteen is nine times two twenty is ten times two so notice that we could cancel a two so the final answer is nine over ten so when multiplying two fractions you could simply multiply across now sometimes you may have to multiply two fractions that contain very large numbers for example 28 over 63 times 56 divided by 35. now you don't want to multiply 28 by 56 because it's going to give you a big number and then you have to simplify that big number which is a lot of extra work the best thing to do is to simplify the numbers now 28 is seven times four sixty-three is seven times nine fifty-six is eight times seven and thirty-five is five times seven so notice that we could cancel a seven on the left and on the right we can also cancel a 7. now there's nothing else that we can cancel so all we got to do is multiply 4 by 8 which is 32 and then 9 times 5 which is 45 and this is the answer notice that we didn't need a calculator for this example so the answer is 32 divided by 45. now what about dividing two fractions what's 36 over 52 divided by 27 over 65 whenever you want to divide two fractions you want to keep in mind this expression keep change flip so what it means is you need to keep the first fraction the same next change division to multiplication and then flip the second fraction so now this problem is like the last problem so instead of multiplying 36 by 65 let's break down 36 into more manageable numbers 36 is 9 times 4. 52 is 13 times 4 65 is 13 times 5 and 27 is 9 times 3 so we could cancel a 13 we can get rid of a 4 and we can get rid of a 9. so the final answer for this problem is 5 divided by 3. so that's just a basic review of fractions things you just have to know if you're going to take algebra trig precal calculus or any other higher level math class so make sure you know these principles now the next thing you need to understand is adding and subtracting like terms for example let's say if you want to simplify this expression five x plus three x squared minus seven x plus four x cubed plus eight x squared now you can't add 5x and 3x squared they're not like terms x and x squared are completely different you can however combine 5x and negative 7x because the five and negative seven they're attached to the same thing that is x so five plus negative seven that's a negative two so you can combine that into negative two x those are like terms we can't add 3x squared with 4x cubed they're not like terms x squared and x cube are different we can however combine 3x squared and 8x squared those are like terms so all you got to do is add 3 plus 8 and that will give you 11. and then the last one you can't combine with anything so that's just 4x cubed so make sure you know when you can combine certain terms and when you can now we mentioned that we cannot add 5 x plus 3x squared but can we multiply 5x times 3x squared so you can't add unlike terms but you can multiply them 5 times 3 is 15. x times x squared is x cubed when you're multiplying variables you need to add the exponents for example x to the fourth times x to the seventh that's going to be x to the four plus seven which is x to the eleventh when dividing you can subtract the exponents so x to the seventh divided by x to the fourth is the same as x to the seven minus four which is x cubed and when you raise one exponent to another let's say x cubed raised to the fourth power you need to multiply the exponents 3 times 4 is 12. so that's x to the 12th now here's something that can help you to understand why that works so let's say if you're multiplying x squared by x cubed i like to think of it this way x squared means that you're multiplying x times x two x variables x cubed means that you're multiplying x times x times x three x variables so in total you really multiply in five x variables and two plus three is five so that's why when you're multiplying variables of the same type you're allowed to add the exponents when you're dividing let's say x to the fifth divided by x cubed x to the fifth is basically five x variables multiplied to each other x cubed is three x variables so you can cancel three on top and three on the bottom which leaves 2 left over on top so x 5 minus 3 is x to the second power which is what we have two x variables left over so when you're dividing you need to subtract the top exponent by the one on the bottom so for example let's say if we have x to the fifth divided by x to the eighth so it's going to be x to the 5 minus 8 which is x to the negative 3. now when you have a negative exponent to make it positive you can take the x variable and move it to the bottom if you do so the negative 3 will now become positive 3. and typically in the algebra course you want to get rid of any negative exponents you want to make all of them positive now if you want to understand why it works this way you could expand the expression x to the fifth is five x variables multiplied to each other and x to the eight is eight x variables multiplied to each other so we can cancel five on top and five on the bottom so notice that it leaves us with three x variables on the bottom that's why the answer is one over x cubed now when you raise in one exponent to another power we said we need to multiply two times three is six now let's understand why x squared raised to the third power means that we're multiplying three x squareds together whenever you have an exponent what it means is repeat multiplication that's what it represents so we're multiplying x squared three times and x squared means that we're multiplying x two times so in total we're multiplying six x variables that's why it's x to the sixth power and so that's why i got to multiply two and three together now let's look at another example so let's say if we want to multiply x to the fourth power three times so it's x to the fourth times x to the fourth times x to the fourth so basically each x to the fourth represents four x variables multiplied to each other so what we have is a total of 12 x variables so it's x to the 12. so 4 times 3 is 12. so anytime you raise one exponent to another power you need to multiply those two exponents so here's some examples that you could work on go ahead and simplify these problems let's say if we have 5x cubed times 7 x to the fourth and then 28 x to the fifth divided by let's say 40 x squared and 3x to the fourth raised to the second power let's start with the first example 5 times 7 is 35 x cubed times x to the fourth so we need to add the exponents 3 plus 4 is 7. so the answer is 35 x to the 7th now let's move on to the second example 28 is seven times four forty is ten times four now x to the fifth i'm gonna break it down into x cube and x squared because two plus three is five so i can cancel a four and i can cancel x squared so the final answer is going to be seven x cubed divided by 10. now in the next example i need to multiply the exponents 3 is the same as 3 to the first power so 1 times 2 is 2. so i have 3 squared and then 4 times two is eight so this is going to be x to the eighth power three squared is three times three which is nine so the final answer is nine x to the eighth power now let's try some other similar examples try these 30 x to the fifth divided by 48 x to the ninth power and then four x cubed y squared multiplied by three x cubed y to the fifth and finally four x to the fourth y to the fifth raised to the third power so take a minute and work on these examples go ahead and pause the video and once you finish it unpause it and check your work so in the first example i'm going to break down 30 into 6 times 5 and 48 is 6 times 8. x to the knife i'm going to write it as x to the fifth times x to the fourth because 5 plus 4 is 9. and now i'm going to cancel so i can cancel a six and x to the fifth so the final answer is five over eight x to the fourth power now for the next example i need to multiply four by three which is twelve and then x cubed times x cubed 3 plus 3 is 6 and then y squared times y to the fifth which is y to the seventh and so this is the answer 12 x to the 6th y to the 7th now for the last example i need to multiply every exponent by three so one times three is three four times three is twelve five times three is fifteen four to the third is four times four times four four times four is sixteen sixteen times four is sixty-four so the solution is sixty-four x to the twelfth power times y to the fifteenth power now it's important to note that anything raised to the zero power is always one so three x y squared raised to zero is just going to be one now if you have a negative outside of it then this is going to be a negative one now let's say if you're given these three problems i want to make sure you understand the difference and what to do if you ever see this on the test what is the negative 3 squared when the negative is not part of a parenthesis when the negative is not inside of parentheses the two only affects the three not the negative sign so what this means is you have negative and then you're multiplying two threes together three times three three times three is nine so your final answer is just negative nine if you type this in exactly the way you see it in the scientific calculator it's going to give you negative nine now let's look at the next example the negative sign is now inside the parentheses so it's affected by the two so there's two of them we'll multiply negative three by negative three which is positive nine again if you were to type this in exactly the way you see it in a calculator it's going to tell you it's positive 9. now the last example has one negative on the outside and on the inside of multiplying two negative threes together so when you're multiplying three negatives overall it's negative negative three times negative three is positive nine and then negative times positive nine overall is negative nine let's say if you were to see something that looks like this 3x times 5x minus 4 and you're asked to simplify this problem what would you do in this case all you can do is distribute that is to multiply three x to five x and also multiply three x by negative four that's the process of distribution three x times five x three times five is fifteen x times x is x squared and then 3x times negative 4 is negative 12x so that's all you could do in this problem whenever you multiply in one term which is the 3x by two terms then you're distributing or if you're multiplying one term by three terms you're distributed a single term is a monomial two terms is a binomial three terms a trinomial if you have many terms a polynomial so this here is a binomial because there are two terms one two now what if we wanted to let's say we want to multiply a binomial two x plus three by another binomial what should we do in this case we need to foil perhaps you heard of this expression the f stands for first you want to multiply the first two terms and then you want to multiply the stuff on the outside 2x by negative 5 and then the stuff on the inside and then finally the last terms but let's do an order so first let's multiply 2x by 3x so that's going to be 6x squared and then let's multiply the terms on the outside 2x times negative 5 which is negative 10x and then let's multiply the two terms on the inside three times three x which is positive nine x and then finally the last terms three times negative five which is negative fifteen now we need to simplify this algebraic expression so we need to combine like terms negative 10x plus 9x that's going to be negative 10 plus 9 which is like 9 minus 10 that's negative 1. negative 1x and negative x are the same so this is the answer it's 6 x squared minus x minus 15. what about this example what's 3x minus 4 raised to the second power what do you think we need to do in this problem well keep in mind whenever you see an exponent it represents repeat multiplication so what we really have is 3x minus 4 times itself two times so once again we need to foil it three x times three x is nine x squared and then three x times negative four that's negative 12x negative 4 times 3x is also negative 12x and finally negative 4 times negative 4 is positive 16. now just like the last example we're going to combine like terms negative 12x plus negative 12x that's negative 24x and so this is the answer 9x squared minus 24x plus 16. now what about multiplying a binomial by a trinomial go ahead and try this problem now when it gets complicated like this i like to line things up so let me show you 2x times 4x squared that's 8 x cubed so you gotta add one plus two which is three next we have two x times negative three x that's going to be negative six x squared and finally two x times 6 is 12x next we have 5 times 4x squared which is 20x squared i'm going to put it below the negative 6x squared and then 5 times negative 3x that's negative 15x and lastly 5 times 6 which is 30. now the reason why i like to align it like this is i could easily identify the like terms negative 6x squared and 20x squared those are like terms 8x cubed there's nothing else like it that i can combine it with so i'm just going to bring it down negative 6 plus 20 that's positive 14. 12 and negative 15 that adds up to negative 3. so this is the answer 8x cubed plus 14x squared minus 3x plus 30. here's another example 3x squared minus 2x plus 4 times 4x squared plus 5x minus 6. so in this example we're multiplying a trinomial by a trinomial so initially before we combine like terms we should have nine terms because three times three is nine in the last example before we combine like terms we had six terms because two times three is six so let's multiply 3x squared by 4x squared so that's going to be 12 x to the fourth next let's multiply 3x squared by 5x and so that's going to be 15 x cubed and then we're going to multiply 3x squared by negative 6 which is just negative 18 x squared now you got to be careful that you don't make a mistake because the longer the problem the easier it is to make a single mistake which can mess up the whole problem so just you need to double check your work when doing problems like this negative 2x times 4x squared that's negative 8x cubed so i'm going to line it up just to avoid mistakes and then negative two x times five x that's going to be negative 10 x squared next we have negative two x times negative six which is positive twelve x now let's move on to the next part 4 times 4x squared that's positive 16x squared and then 4 times 5x that's 20x and finally 4 times negative 6 which is negative twenty four so as you can see we have a total of nine terms before combining like terms three six nine now we can't combine 12x to the fourth of anything so let's bring it down 15 plus negative 8 that's positive 7 and then negative 10 plus 16 that's positive 6 and negative 18 plus 6 that's going to be negative 12. 12 plus 20 is 32 and so this is the answer 12x to the fourth plus 7x cubed minus 12x squared plus 32x minus 24. now let's talk about solving basic equations you need to master skill especially if you're going to go into an algebra course let's start with this one x plus 8 is equal to 15. x is basically a variable in which you currently don't know what the value of x is a number it represents a specific number so what number plus 8 is equal to 15. well it turns out that seven plus eight is fifteen so therefore x has to be seven now how do we show our work how do we show that x is equal to whenever you need to solve an equation whenever you need to find a value of a variable the best way to do that is to get that variable by itself on any side of the equation so basically if we can move the 8 from the left side to the right side we can get the x variable by itself on the left side and the value that's on the right is going to be equal to the value of x and that's basically what you're doing when you're solving an equation you want to find the value of a variable now the opposite of addition is subtraction so to get rid of the 8 on the left we need to subtract both sides by 8. 8 plus negative 8 is 0. so now what we have left over is x is equal to 15 minus 8 which is 7. and so that's how you could find the value of x let's look at another example x minus 4 is equal to 12. so once again we need to get x by itself on the left side or on the right side to find its value so let's keep it on the left but let's get rid of the negative 4 on the left the opposite of subtraction is addition so we need to add 4 to both sides 12 plus 4 is 16 and you could check the work 16 minus 4 is equal to 12. so x has to be 16. now what about this example if 3x is equal to 15 what is the value of x so how can we separate the 3 from the x how can we get rid of the 3 on the left side well right now the 3 is multiplied to x so if you want to get rid of it from the left side perform the opposite operation of multiplication the opposite of multiplication is division so what you want to do is divide both sides by three three divided by three is one so on the left we're gonna have one x which is the same as x fifteen divided by 3 is 5 and that is the answer x is equal to 5. now if x divided by 6 is equal to 4 what is the value of x the opposite of division is multiplication so in this example we want to multiply both sides by six six divided by six is one so those two numbers will cancel and on the right we have four times 6 which is 24 and so that's the value of x now what about this example two-thirds of x is equal to eight what is the value of x so what you can do is you can multiply by the reciprocal of the fraction so let's multiply the left and the right sides by 3 divided by 2. on the left side the threes will cancel and the twos will cancel so all we have left over is x eight is the same as eight divided by one and you know how to multiply fractions you gotta multiply across so eight times three is twenty four and one times two is two and twenty four divided by two is twelve and this is the answer now let's move on to more difficult examples so sometimes you may need to solve a multi-step equation like this one so if 3x plus 5 is equal to 11 what is the value of x now you don't want to divide everything by three if you do it's just it's not going to be nice instead you want to subtract both sides by 5 first if you do it this way it's going to be a lot easier and with practice you'll get the hang of it eleven minus five is six and now what we can do is divide both sides by three so x is equal to 2. the best way to learn algebra is to make sure to work through the most of the homework problems that can be found in your textbook the more problems you do the better you'll become with math so it's based on the effort that you put in the more effort you put in the more you get out from it try this example let's say 2x minus 7 is equal to 3. find the value of x so we're not going to worry about the two right now let's get rid of the seven on the left side so let's add seven to both sides three plus seven is ten and so two x is equal to ten now once you get to this part you can separate the two and the x by division and ten divided by two is five so this is the solution x is equal to five and you could plug it back in to see if the answer is correct so two times five minus seven is that equal to three two times five is ten and ten minus seven is three so the equation is balanced now let's move on to an example where we have variables on both sides of the equation so let's say that 4x plus 3 is equal to 6x minus 15. now what do you think we need to do in this example your goal is to get all of the x variables on one side and every number that doesn't have an x variable on the opposite side so because six is larger than four i'm going to subtract both sides by four i don't want to deal with negative numbers if i don't have to now at the same time we can add 15 to both sides so all of the variables i'm moving it to the right side i no longer have an x variable on the left side and all of the constants the numbers without a variable is all on the left side 3 plus 15 is 18 6x minus 4x is 2x so once i get to this part of the equation all i need to do is divide both sides by 2. 18 divided by 2 is 9. so as you can see solving equations linear equations is not very difficult you can learn this you can do this now what would you do if you have a linear equation that contains parentheses and variables on both sides so if you see this don't let it concern you you can do this the first thing you should do is distribute so let's multiply 3 by 2x so this is equal to 6x and then multiply 3 by negative 4 which is negative 12. now on the right side distribute the 5. 5 times 3x is 15x and then 5 times 2 that's 10. and let's not forget the negative 3 that we have on the right side so at this point combine like terms if there are any on the right side we can combine 10 minus 3 which is 7. so now we're at the point at the start of the last example 15 is larger than 6 so i'm going to subtract both sides by 6x so i will no longer have a variable on the left side now i want to move all my constants to the left side because the x variable is now on the right side so therefore i need to get rid of the seven so i'm going to subtract both sides by seven so all i have on the right side is 15x minus 6x which is 9x on the left side negative 12 minus 7 which is negative 19. now all you could do at this point is divide by 9. so x is equal to a fraction negative 19 divided by 9. now you may get examples where you have to solve equations that contain fractions so consider this problem 2 two-thirds x plus five is equal to eight what is the value of x now what i like to do is i like to get rid of the fraction from the beginning there's many different ways of going about solving this equation but this is just the method that i prefer i'm going to multiply everything by three by the denominator of this fraction so three times two-thirds the threes will cancel leaving behind simply 2x and then i'm going to multiply 3 by 5 which is 15 and finally 3 times 8 which is 24 and so now i have an equation that is a lot easier to deal with so next i'm going to subtract both sides by 15 and 24 minus 15 that's nine and then divide both sides by two so x is equal to 9 over 2. since the numerator is greater than the denominator we have what is known as an improper fraction so what you can do is you can turn it to a mixed number 9 is basically eight plus one eight divided by two is four and four plus one half is four and one halves as a mixed number so x is basically equal to four point five as a decimal let's try another example three-fourths x minus two-thirds is equal to twelve calculate the value of x now this time there are two fractions in this equation so should we multiply both sides by three or by four if you want to get rid of both fractions you should multiply by both numbers three and four three times four is twelve i'm going to leave it as 4 and 3. so if we multiply this fraction 3 4 by 12 what will we get well the 4s will cancel and so we're left with 3 times three which is simply nine so therefore three fourths x times twelve is going to be nine x now we need to multiply two thirds by 12. so as you can see the threes will cancel and we're left with two times four which is eight and 12 times 12 is 144 now we no longer have any fractions so let's finish the problem let's add 8 to both sides 144 plus 8 is 152 and then we could divide both sides by nine so x is 152 divided by nine now let's turn this into a mixed number and this time i'm going to use the long division so how many times does 9 go into 15 9 goes into 15 one time 9 times 1 is 9 9 times 2 is 18 which exceeds 15. now 15 minus 9 is six so now let's bring down the two nine goes into 62 six times nine times seven is 63 so that's too much nine times six is uh 54 and 62 minus 54 is 8. so to convert this into a mixed number 9 goes into 152 16 times that's going to be the number outside of the fraction and then the denominator is going to stay 9 and then what remains the remainder is the numerator of the mixed fraction so it's 16 and 8 over 9. now other times you may have linear equations that contain decimals i think this is going to be easier than the last example so let's try go ahead and calculate the value of x in this problem now notice that every number is rounded or it's at the the tenths place so therefore multiply everything by ten point eight times ten is eight point three times ten is three point five times ten is five and one point four times ten is fourteen and now it's a lot easier to solve it so let's subtract three from both sides and let's subtract five x from both sides 8x minus 5x is 3x and 14 minus 3 is 11. so the last thing we need to do is divide by 3 so x is equal to eleven over three eleven over three is basically nine over three plus two over three and nine divided by three is three so we got three plus two thirds which as a mixed number is three and two thirds or 3.67 as a decimal now let's look at another example 0.34 x plus 0.12 is equal to 0.26 minus point 15 x go ahead and find the value of x so in this example notice that each number is at the hundreds place so to turn everything into a whole number multiply everything by a hundred point three four x times a hundred is thirty four x and then the next number is going to be twelve and then twenty six minus fifteen x so basically you just gotta move the decimal point two units to the right if you're multiplying a number by 100 now let's solve as we've been doing before so let's add 15x to both sides and let's subtract both sides by 12. 34 plus 15 that's 49. 26 minus 12 is 14. so now let's divide both sides by 49. so x is equal to 14 over 49 which we could simplify that fraction 14 is 7 times 2. 49 is 7 times 7. so therefore x is equal to 2 divided by 7. and so that's the solution for this problem what would you do if you have two fractions separated by an equal sign so let's say x divided by five is equal to eight over nine what is the value of x whenever you have two fractions separated by an equal sign you can cross multiply five times eight is forty and x times nine is nine x and then we could divide by 9. so x is equal to 40 over 9. and that's all you got to do if you have two fractions separate by an equal sign so let me give you another example let's say we have 3x plus 2 divided by five is equal to two x minus five time i mean divided by four calculate the value of x so first cross multiply so we have four times three x plus two and that's equal to five times two x minus five now in this example we need to distribute so let's distribute four to three x plus two four times three x is twelve x and four times two is eight five times two x is ten x and five times negative five is negative twenty five and then it's like another problem that we've seen multiple times so let's subtract both sides by 10x and let's subtract both sides by 8. minus 10x is 2x negative 25 minus 8 is negative 33 and then we'll just divide by 2. so x is equal to negative 33 over 2 which is negative 16.5 as a decimal now you need to be familiar with inequalities that's going to be our next topic of discussion go ahead and graph these four inequalities on a number line so x is greater than two so what does that mean how can we show that on a number line if you look at the second example it means that x is equal to or greater than one let's put a zero here let's put a two if you just have the greater than symbol on a number line you need to draw an open circle so x is greater than two we're gonna put an open circle at two and because it's greater than we're gonna shade towards the right because all of the numbers that are larger than 2 or have a higher value than 2 they're towards the right of 2. now for the second example x is equal to or greater than 1. so we're going to have a closed circle to include one but we're still going to shade to the right because x can be greater than one for the third example x is less than negative two so negative two is to the left of zero and because it's not equal to negative two it doesn't include it we need to use a open circle and shade towards the left because it's less than negative two now for the last example x is less than or equal to negative three so this time we're going to use a closed circle but we're still going to shade towards the left now you need to be able to represent these solutions using interval notation for the first example in the upper left x is greater than 2 so the way you can write it is 2 to infinity all the way to the right you have positive infinity to the left negative infinity so it starts from 2 and it ends at infinity if you have an open circle always use parentheses and for infinity symbols you should always use parentheses now for the example in the upper right this time one is included so we need to use a bracket so it's shaded starting from one and then towards infinity so it's from one to infinity so use brackets if one is included now for the third example on the lower left it's shaded starting from negative infinity to negative two so the way you look at it is from left to right and that's the way you write it from left to right so it's going to be negative infinity to negative two and since we have an open circle we're going to use parentheses as opposed to uh brackets now for the last example it starts from negative infinity and it stops at negative three so it includes negative three so we're going to use brackets there now go ahead and solve these two equations five plus three x is greater than let's say 11. actually let's make it 14. and also solve this one two minus 5x is less than or equal to let's say negative eight now i'm going to start with the example on the left so whenever you're solving an inequality you can treat it as if it's an equation view this as if it was an equal sign so you're going to have 3x is greater than 14 minus 9 i mean 14 minus 5 which is 9 and then divide by 3. so x is greater than 3. now once you get this answer represent it in two ways using a number line and using interval notation so because x is greater than three we're going to shade towards the right and it doesn't equal three so we're gonna use an open circle so in interval notation is from three to infinity now for the example on the right we need to subtract two from both sides so negative five x is less than or equal to negative ten now what we need to do is divide both sides by negative five now here's what we got to be careful whenever you multiply or divide by a negative number you need to switch the sign of the inequality so it's going to change from less than or equal to to greater than or equal to negative 10 divided by negative 5 is positive 2. so this is our answer and now let's plot it on a number line so this time we're going to use a closed circle and we're going to shade towards the right because it can be greater than 2. so an interval notation is from 2 to infinity using brackets since 2 is included so that's it for this video if you want to find more videos on algebra you can check out my udemy course or you could check out my channel for videos in my algebra playlist so thanks again for watching now i want to show you one of my algebra courses that might be useful to you if you ever need it so go to udemy.com now in the search box just type in algebra and it should come up so it's the one with the image with the black background so if you select that option and if you decide to go to course content you can see what's in this particular course so the first section basic arithmetic for those of you who want to focus on addition subtraction multiplication and division and it has a video quiz at the end it's a multiple choice video quiz you can pause it work on the problems and see the solutions it covers long division multiplying two large numbers and things like that the next tutorial is on fractions adding subtracting fractions multiplying dividing fractions converting fractions into decimals and so forth so you can also take a look at that next solve the linear equations which we covered and just more examples if you need more help with that the next topic order of operations which is also useful uh graphing linear equations you need to know how to calculate the slope needs to be familiar with the slope intercept form standard form and just how to tell if lines are parallel perpendicular and so forth and there's a quiz that goes with that as well the next topic is on inequalities and absolute value expressions which are also seen a typical algebra course and then we have polynomials and that's a long section and then factoring you just that's another topic you need to master and then system of equations you can solve it by elimination substitution there's also word problems as well sometimes you got to solve equations with three variables x y and z so that could be helpful next quadratic equations how to use the quadratic formula how to graph them how to convert between standard and vertex form and then you have rational expressions and radical expressions solving radical equations simplifying it things like that and every section has a quiz so you can always review what you've learned if you have a test the next day so here we have complex imaginary numbers you need to know how to simplify those exponential functions logs i have a lot of videos on logs and then this is just functions in general vertical line tests horizontal line tests how to tell for functions even or odd and then conic sections graphene circles hyperbolas ellipses parabolas and things like that there's two video quizzes because it's actually a long section and finally arithmetic and geometric sequences and series so that's my algebra course if you want to take a look at it and uh let me know what you think you