00:0-1 | Okay , Here's a video that's going to review all | |

00:01 | main concepts you learned in grade nine . Math . | |

00:03 | In under 60 minutes , you're gonna learn all grade | |

00:06 | nine and under an hour . This is a great | |

00:07 | video if you're doing exam review , or if you're | |

00:10 | about to take the course and just want to know | |

00:11 | what you're going to learn now in this video , | |

00:13 | I'm not going to go in depth into any of | |

00:15 | the topics . I'm going to go through a couple | |

00:17 | examples of each topic . But if you want in | |

00:19 | depth explanations and proofs of each topic going to Jensen | |

00:22 | math dot c A . And there's video lessons for | |

00:24 | each topic there . So I'm going to divide this | |

00:27 | video into three parts based on the three main units | |

00:29 | you do in grade nine . So the first part | |

00:31 | is going to be on algebra and algebra . You | |

00:35 | learn about polynomial is you learn how to collect like | |

00:37 | terms you learn distributive property , learn experiment laws and | |

00:40 | the next part of solving equations . So I'm gonna | |

00:42 | go through that part first and watch the second video | |

00:45 | . If you want to learn about linear relations and | |

00:46 | the third video , if you want to review the | |

00:48 | geometry section . So watch all three videos and you'll | |

00:52 | have reviewed all of the main concepts of Grade nine | |

00:54 | math in under an hour . So let's get started | |

00:57 | with the first topic right away . I'm going to | |

00:59 | go quickly just so I can get this done in | |

01:01 | under an hour . So the first main topic you | |

01:04 | would have learned in grade nine math is exporting vase | |

01:06 | . So to do export laws . First main thing | |

01:08 | you have to remember , you have to know what | |

01:10 | power is . So let's say we have five squared | |

01:12 | . We have to know that that means five times | |

01:15 | five , right ? Um , this is the base | |

01:18 | . Five is the base , two is the exponent | |

01:20 | , and this is a power . So the exponent | |

01:22 | tells you how many of the base So how many | |

01:24 | factors of the base are being multiplied together ? So | |

01:27 | there are some exponent laws you would have learned the | |

01:28 | product of powers rule . So if you have two | |

01:30 | powers being multiplied together , the dot means multiply . | |

01:33 | So to power is being multiplied together , have the | |

01:35 | same base . That's important . Base has to be | |

01:37 | the same . There's a rule . You keep the | |

01:39 | base the same , and you add the exponents that | |

01:42 | equals X to the A plus piece . You can | |

01:44 | rewrite as a single power by keeping the base the | |

01:46 | same . Adding the exponents . What if you have | |

01:48 | two powers for the same base being divided by each | |

01:51 | other ? There's a similar rule for that one that's | |

01:54 | similar rule . You keep the base the same , | |

01:56 | and this time you subtract the exponents . That's actually | |

01:59 | a minus . B . You would have learned the | |

02:01 | power of the power rule . So if you have | |

02:02 | an exploring on top of an exponent , keep the | |

02:05 | base and you multiply the exponents X to the A | |

02:08 | Times B . What if you have a power where | |

02:10 | the base is the quotient to the power of a | |

02:12 | quotient ? You have to remember that this experiment gets | |

02:14 | put onto the numerator and the denominator , so this | |

02:17 | is equal to age . The X over B to | |

02:21 | the power of X . What if we have a | |

02:23 | power where the base is a product of numbers or | |

02:25 | variables , so this exponent outside of the brackets gets | |

02:28 | put onto each of the factors inside the brackets , | |

02:32 | so this would be equal to a to the X | |

02:35 | Times B to the X and a couple more rules | |

02:41 | you would have learned anything to the power of zero | |

02:43 | is equal to one and the negative exponent rule . | |

02:46 | If you have power with the negative exponent , you | |

02:48 | can't leave . A final answer with the negative exponent | |

02:50 | in this section is what you have to do is | |

02:51 | rewrite this the positive exponent by running the reciprocal of | |

02:54 | it . So it would be one over X to | |

02:57 | the power of positive A . So I'm gonna go | |

02:59 | with you . A couple examples that review all of | |

03:01 | these rules that these are the main exponent rules would | |

03:03 | have learned in grade nine . So let's practice a | |

03:06 | few of them without variables first . Then I'll throw | |

03:08 | some variables into the mix . So here we have | |

03:10 | powers that all have the same base right there , | |

03:12 | all the base of four . The first two are | |

03:14 | being multiplied together . So what I can do is | |

03:16 | rewrite as a single power by adding the experience three | |

03:19 | plus 58 And then I have to divide that by | |

03:22 | four squared so I can rewrite that Since they have | |

03:24 | the same base and they're being divided . I keep | |

03:26 | the base the same . And I subtract the exponents | |

03:29 | . So I end up with four to the power | |

03:30 | of six . And then after I've written it as | |

03:32 | a single power , I can evaluate it . Four | |

03:34 | to the power of six is 4000 96 . Number | |

03:38 | four of the power of six means four times , | |

03:40 | four times , four times , four times , four | |

03:41 | times four . It doesn't mean 4 , 10 , | |

03:43 | 6 . Okay , power of a quotation . And | |

03:45 | we have power of the power here as well . | |

03:48 | So let's start off with the power of power rule | |

03:50 | . So I have an exponent on top of an | |

03:53 | exponent . I can rewrite this , um , by | |

03:57 | multiplying these powers together two times three is six and | |

04:02 | notice I put the negative sign into the numerator . | |

04:03 | I could have put it into the denominator if I | |

04:05 | wanted to , but don't put it into both . | |

04:06 | I decided to put it into the numerator to make | |

04:08 | it easier for me here . So here I have | |

04:10 | negative to over three to the power of six . | |

04:12 | So what I have now is power of a quotient | |

04:14 | . So remember , for power of the quotient the | |

04:16 | exponent . If the base is a quotation , the | |

04:18 | exponent of the power goes onto the numerator and the | |

04:22 | denominator . So this equals negative two to the 6/3 | |

04:26 | to the six , and then I have to let | |

04:29 | me move this down . And then I have to | |

04:32 | evaluate each of those powers . So negative two to | |

04:36 | the power of six is 64 and three to the | |

04:39 | power of six is 7 29 . So that's fully | |

04:43 | simplified . I can't reduce that fraction any further . | |

04:46 | Let's look at these two here . So I have | |

04:49 | a product of powers here and the powers have the | |

04:51 | same base . So I can use my expletive law | |

04:52 | to simplify by keeping the same base and adding the | |

04:55 | exponents four plus negative . Seven . Be careful with | |

04:57 | your integers . Four plus negative seven negative three . | |

05:00 | But I can't leave that as my final answer with | |

05:02 | the negative exponents . What I have to do is | |

05:04 | rewrite that with a positive exponent by running the reciprocal | |

05:07 | of it . So 1/5 to the power of positive | |

05:10 | three . And now I can evaluate five to the | |

05:12 | power of three that gives me 1 25 . So | |

05:16 | 1/1 25 is my final answer there . How about | |

05:19 | here . I have quotient of powers . So what | |

05:22 | I have to do ? I can rewrite it as | |

05:24 | a single power by keeping the base and subtracting the | |

05:27 | exponents seven minus 70 Remember , anything to the explosion | |

05:30 | of zero is equal to one , and you should | |

05:32 | notice up here when you have something divided by itself | |

05:35 | . That answer is always going to be one as | |

05:37 | well . So I think of it . Either way | |

05:39 | , let's throw some variables into the mix . So | |

05:41 | I have a product of powers . Here I have | |

05:44 | these three powers being multiplied together . They're all the | |

05:46 | same base , so I can rewrite as a single | |

05:48 | power by keeping the same base and adding the exponents | |

05:51 | three plus four plus five . That's 12 8 of | |

05:54 | the 12 . I can't evaluate it . Evaluate that | |

05:56 | since it is a variable sweet , leave it like | |

05:58 | that . There's our final answer . What we have | |

06:01 | here is power of a product , so the base | |

06:04 | of the power is a product of four x square | |

06:06 | , N y to the five . So what I | |

06:08 | have to do is put this outside exponents onto each | |

06:12 | of the factors in the product onto the four onto | |

06:14 | the X squared and on why to the five . | |

06:16 | So I have to do four cubed . I'll put | |

06:19 | in brackets just so it looks nicer here . I | |

06:21 | have to do X squared , cubed , and I | |

06:24 | have to do . Why to the five cubed right | |

06:27 | , This outside exponent goes on to each of the | |

06:29 | factors of the product of the base . And now | |

06:32 | I can evaluate four Cubed is 64 x to the | |

06:36 | two to the three . Well , that's the power | |

06:37 | of the power . I don't have to multiply the | |

06:39 | experience there . So that's X to the sixth and | |

06:41 | why to the five to the three power of the | |

06:43 | power rule again . Why ? To the 15 and | |

06:46 | that's done here . I have product of some powers | |

06:50 | and some coefficients here , so let's start off with | |

06:53 | the five times four . Always start with your coefficients | |

06:56 | . We can multiply five times four , just as | |

06:58 | you always would . It's 20 . Now you look | |

07:01 | for powers that have the same base . I have | |

07:03 | an M two . The five and an M squared | |

07:05 | does have the same base , so I know when | |

07:06 | I multiply powers to have the same base , I | |

07:09 | keep the base . The same . And I add | |

07:10 | the exponents five plus two is seven . So I | |

07:13 | m to the seven . Also , I have two | |

07:15 | powers of em that are being multiplied , so I | |

07:19 | can simplify that by writing it as a single power | |

07:21 | by adding the exponents and keep in mind , this | |

07:23 | end has an exponent of one , even though you | |

07:25 | don't see an exponent there . So one plus four | |

07:28 | is five , and that is fully simplified there . | |

07:31 | That question is done . Let's look at a quotation | |

07:35 | of powers here . Um , same thing . Start | |

07:38 | with the coefficients . Start with your 36 divided by | |

07:40 | 27 divide 36 by 27 exactly as you always would | |

07:44 | . But we don't want a decimal answer . We | |

07:45 | just want to reduce it right . 27 doesn't go | |

07:47 | into 36 evenly , but I can reduce that fraction | |

07:50 | by fighting number that goes into both of them evenly | |

07:53 | . And in fact , nine goes into both 36 | |

07:55 | 27 9 goes into 36 4 times and it goes | |

07:59 | into 27 3 times so I can reduce 36/27 to | |

08:03 | 4/3 . Now I look at my variables , so | |

08:05 | I have some powers , but with the same basis | |

08:08 | . I have an X cubed in an extra six | |

08:10 | that are being divided by each other . So I | |

08:12 | know I can simplify that writing as a single power | |

08:15 | if I subtract the exponents three minus six is negative | |

08:19 | . Three . And make sure you put your quotations | |

08:21 | into the numerator Here will take care of that negative | |

08:23 | in a second . Next . We also have two | |

08:25 | powers . Have a base of Why ? Why did | |

08:27 | the nine divided by y to the four reread ? | |

08:29 | It's a single power by keeping the base and subtracting | |

08:31 | the exponents . Nine . Minus four is five . | |

08:33 | Remember , always put your questions into the numerator . | |

08:36 | Now we have to do something with this answer . | |

08:37 | We can't leave our answer Where The power that as | |

08:40 | a negative exponent , this X to the negative three | |

08:43 | is a power with the negative exponents . We can | |

08:45 | take care of that negative by regular reciprocal of it | |

08:47 | . And what's going to happen is this X to | |

08:49 | the negative three . Not before just the excellent negative | |

08:52 | three . Just that power . If we bring it | |

08:55 | into the denominator , it'll make the exponent positive . | |

08:58 | So what ? We have for our final answer . | |

09:00 | The four and the white of the five stay in | |

09:02 | the numerator . The three states in the denominator , | |

09:04 | and we bring the power of extra negative three to | |

09:06 | the denominator , and it makes it a power of | |

09:09 | positive three . So there's my final answer there . | |

09:13 | Let's do one more . If you can do this | |

09:14 | example , you can be confident that you understand all | |

09:18 | of the exponent laws . So for this example , | |

09:20 | start with just the numerator . Forget about the time | |

09:23 | there . For now , let's look at just the | |

09:24 | numerator and simplify that I have an eight times before | |

09:27 | . That's 32 . I have no hope for powers | |

09:30 | to the same base of a B cubed times of | |

09:32 | B to the one . Keep the base . Add | |

09:35 | the exponents . Remember product rule when multiplying power , | |

09:38 | to say , basically , add the exponents . And | |

09:40 | now I have a D to the one times a | |

09:41 | D to the two that's D to the three . | |

09:44 | If we look in the denominator , let's just leave | |

09:47 | this to out front for now . And let's just | |

09:51 | do this power of the product rule here with this | |

09:53 | expletive , too , has to go on each factor | |

09:55 | of the base that's still on the to the B | |

09:58 | and the deeds . We have to square the two | |

10:00 | , which makes it a four . We have to | |

10:01 | square the B , which makes it a B squared | |

10:03 | and square the d , which makes it a D | |

10:05 | squared . So now I have two times four b | |

10:07 | square d square , So I can simplify that , | |

10:10 | um , two times four and make it an eight | |

10:13 | . So be to the four . Leave the numerator | |

10:15 | for now . Two times 48 and then I'd be | |

10:19 | squared d squared and last up on Stewart Dividing 38 | |

10:24 | 32 Divided by eight Divide Those just like you would | |

10:27 | divide annual numbers . Um , 32 divided by it | |

10:30 | goes into 32 4 times . So my answer for | |

10:34 | now , do you be to the four Divided the | |

10:36 | beat of the two When dividing powers the same base | |

10:38 | , keep the base . Subtract the exponents D to | |

10:41 | the three divided by D to the to once again | |

10:43 | keep the base Subtract the experience is the D to | |

10:45 | the one . If it's a one , you don't | |

10:47 | have to write the one . All right , that's | |

10:49 | it for exponent . Law Review . Let's move on | |

10:52 | to polynomial is quickly , so basically , you have | |

10:54 | to remember what a term is . A term is | |

10:56 | an expression form of the product of numbers and variables | |

10:59 | . So , for example , two X that's a | |

11:01 | term . It's a product of a number two and | |

11:04 | a variable X and what a polynomial is , which | |

11:07 | an expression consisting of one or more of these terms | |

11:10 | that are connected by addition or subtraction operators . So | |

11:14 | , for example , I could have the polynomial four | |

11:17 | x squared plus three x plus one . That's a | |

11:20 | polynomial where we have three terms . Term . 123 | |

11:25 | They're separated by addition or subtraction signs . Now we | |

11:29 | can classify a polynomial based on how many terms it | |

11:31 | has by name . If a polynomial only has one | |

11:33 | term like this two . X up here , we | |

11:35 | call that a mono meal . If a polynomial has | |

11:38 | two terms , we call it a binomial . If | |

11:41 | it has three terms , we call it a try | |

11:43 | . No meal , and if it has anything more | |

11:47 | than that , there's no special name for it . | |

11:48 | Like that's four terms . We just simply call it | |

11:50 | a four term polynomial . If it had five , | |

11:54 | we would call it a five term polynomial and so | |

11:57 | on . So one more thing you'll have to be | |

12:00 | able to do is state the degree of a term | |

12:02 | and the degree of the polynomial . Now , to | |

12:04 | state the degree of a term by looking at one | |

12:06 | individual term , you can state the degree of it | |

12:08 | just by adding up . So finding the some of | |

12:11 | the exponents on all the variables in that term . | |

12:16 | So if we look at this first example here three | |

12:18 | x squared , why , this is one term . | |

12:20 | So this is a mono meal . There's nothing else | |

12:22 | added or subtracted from it . It's one term has | |

12:26 | two variables . So to find the degree of this | |

12:28 | term , we add the exponents on the variables . | |

12:30 | Humongous . Why has a one so on the X | |

12:32 | and on the Y , we add the exponents two | |

12:34 | plus one is three . So this this right here | |

12:37 | this term right here is degree three . If we | |

12:41 | look at this next example here , what we have | |

12:43 | here is three different terms being added together . So | |

12:47 | we call this a try . No meal . We | |

12:49 | could find the degree of each term like this first | |

12:52 | term . Here is degree five . This first term | |

12:55 | . Here's degree four because that has an explorer of | |

12:57 | one . And this third term is degree six . | |

13:00 | So if we want to figure out the degree of | |

13:01 | the entire polynomial , all we have to do is | |

13:04 | figure out the degree of the highest degree term in | |

13:08 | this polynomial . We don't add all these degrees together | |

13:11 | . We just pick the term that has the highest | |

13:13 | degree . So in this case , it will be | |

13:16 | the third term . It's degree six . So what | |

13:18 | we say is the entire polynomial is degrees six . | |

13:20 | The degree of the polynomial is equal to the degree | |

13:22 | of the highest degree term in the polynomial . Here | |

13:25 | we have a binomial right . Two terms 12 separated | |

13:30 | by a subtraction sign . The first term is degree | |

13:34 | seven , right , one plus five plus one is | |

13:36 | seven . This term is degree six . So to | |

13:39 | find the degree of the entire polynomial , it's equal | |

13:42 | to the degree of the highest degree term . This | |

13:45 | term is the higher degree . So the degree of | |

13:47 | the entire polynomial is seven . Let's look at collecting | |

13:51 | like terms now . So , first of all , | |

13:53 | what are like terms like terms are terms that have | |

13:56 | the exact same variables with the exact same exponents , | |

13:59 | for example , these two terms . They have the | |

14:01 | exact same variables . They both have an X and | |

14:03 | Y , and on those variables are the exact same | |

14:05 | exponents on the excess of two . And on the | |

14:08 | Y is a one on both of them . So | |

14:10 | those are like terms . The fact that the coefficients | |

14:12 | are different do not make them not like terms . | |

14:15 | The coefficients don't matter when deciding if they're like terms | |

14:17 | or not . You just look at the variables and | |

14:19 | the exponents . So if we look at these two | |

14:22 | here , these are not like terms . Why ? | |

14:24 | Because this one has a Y to the power of | |

14:26 | one . This one has a Y to the power | |

14:28 | of two , so they don't have the same variables | |

14:29 | with the same exponents . So they are not like | |

14:31 | terms . Why do we have to be able to | |

14:33 | know what like terms are ? Because we can . | |

14:35 | We can collect like terms together . For example , | |

14:38 | here I have four terms . So when that are | |

14:43 | being added to subtract from each other when adding or | |

14:45 | subtracting terms , we can collect them together by adding | |

14:49 | or subtracting the coefficients only and keeping the variable of | |

14:53 | the same . This is different than exponent laws with | |

14:56 | exponents . Laws were using them When we're multiplying , | |

14:59 | powers are dividing powers , but now we're going to | |

15:01 | be adding or subtracting terms and we can collect them | |

15:04 | . We can collect like , terms together . So | |

15:06 | what we want to first do is group the terms | |

15:08 | their like terms together and make sure the sign that | |

15:10 | is to the left of the term stays with it | |

15:12 | . So , for example , I have a negative | |

15:14 | two X and a negative five x . Those are | |

15:16 | light terms . So let's write those beside each other | |

15:19 | , keeping the signs letter to the left of them | |

15:21 | . And I also have a positive seven wine , | |

15:23 | a negative nine . While let's write those beside each | |

15:25 | other because those are like terms now I can collect | |

15:28 | them together . I can collect the negative two X | |

15:31 | minus five x together because they're like terms by just | |

15:35 | adding or subtracting the coefficients only so negative . Two | |

15:38 | minus five . That's negative . Seven . And then | |

15:40 | you keep the variable exactly the same . Don't change | |

15:43 | the explosion on the variable it all it stays exactly | |

15:45 | the same . Don't get this confused with exponents laws | |

15:48 | . Now let's look at this group of like terms | |

15:50 | . The seven y minus nine y seven minus nine | |

15:53 | is negative . Two . So I write negative , | |

15:54 | too . Why ? So that is fully complete . | |

15:58 | I can't collect these two terms together because they are | |

16:00 | not like terms , so that expression is fully simplified | |

16:03 | . Let's look at the next expression Here it's a | |

16:05 | little bit longer . Find the groups of , like | |

16:07 | terms . I have a three X squared y and | |

16:11 | an eight x squared y Those both have the exact | |

16:13 | same variables for the same exponents . So I'm going | |

16:15 | to write those beside each other . Three x squared | |

16:17 | y eight x squared y I have a four y | |

16:23 | and a negative one y . So Alright , my | |

16:25 | positive for why am I negative one y remember the | |

16:28 | coefficient is one if you don't see it , and | |

16:30 | I also have to constant terms . Concentrate means in | |

16:32 | turn , without a variable . Those are like terms | |

16:34 | . I have a positive seven and a positive 80 | |

16:37 | right . Those beside each other . Now collect your | |

16:39 | like terms So three X squared y plus X squared | |

16:42 | Y Just add the coefficients . Only 11 x squared | |

16:46 | y Don't change the exponents on the variables at all | |

16:49 | . When you're adding or subtracting light terms together , | |

16:51 | I have a positive four y minus one way . | |

16:54 | That's positive . Three y and I have positive . | |

16:56 | Seven plus 80 . That's positive . 87 . That | |

17:00 | expression is fully simplified . None of these three things | |

17:03 | are like terms with each other , so I can't | |

17:04 | collect them together . Don't try and go any further | |

17:06 | than you can and notice the order I wrote these | |

17:09 | in . You should do it in this order . | |

17:11 | The highest degree term goes first and then it goes | |

17:14 | in descending order . This term is degree three , | |

17:16 | right ? Two plus one is three . This terms | |

17:18 | degree one and a constant terms degree zero . So | |

17:20 | the degree should go in descending order . Yeah , | |

17:24 | Okay , let's look at distributive property . So basically | |

17:26 | for multiplying a mono meal by a polynomial . In | |

17:29 | this case of binomial , this is how you do | |

17:31 | it . Everything inside the brackets gets multiplied by the | |

17:34 | term out front of the brackets and that gets rid | |

17:36 | of the bracket . So if I want to do | |

17:37 | a Times X plus , why I have to do | |

17:39 | a Times X right here and I have to do | |

17:43 | eight times why here . So , for example , | |

17:45 | if I have five times four X plus two , | |

17:48 | I need to multiply the four X and the to | |

17:50 | both of them by the five that's out front . | |

17:52 | So five times four x 20 x five times two | |

17:55 | is positive . 10 . There's my solution . Those | |

17:57 | can't be collected together because they're not like terms . | |

18:00 | So let's practice distributive property . So here I'm doing | |

18:02 | a mono meal multiplied by a try . No meal | |

18:05 | . So everything in the brackets needs to be multiplied | |

18:07 | by the term outfront . Be careful with your signs | |

18:11 | and then that will get rid of the bracket . | |

18:13 | So I have to do negative three times two X | |

18:15 | squared . Well , negative . Three times two is | |

18:16 | negative . Six . So I'm negative . Six X | |

18:18 | squared . I have two negative three times negative . | |

18:21 | Five x Don't forget . This sign belongs to this | |

18:24 | five x So negative times negative is positive . 15 | |

18:28 | x and after your negative three times positive for that's | |

18:31 | negative 12 . And I can't collect any of these | |

18:34 | three terms together because they are not like terms . | |

18:37 | How about here ? I'm going to have to multiply | |

18:40 | the X plus three by the three out front . | |

18:43 | I'm going to have to multiply this X in this | |

18:45 | one by the two out front that will get rid | |

18:47 | of the brackets So three times x is three x | |

18:49 | three times three is positive 92 times X positive two | |

18:54 | x two times one positive to now I can collect | |

18:57 | like terms here because I have a three x and | |

18:59 | two x Put those together . That's five X and | |

19:02 | I have a nine and the to put those together | |

19:03 | , that's positive . 11 . Let's do one more | |

19:07 | example here for collecting like terms . I have four | |

19:11 | K times K and times negative three . Now keep | |

19:14 | in mind this K inside the brackets . Think that | |

19:16 | is a one K if you want . So when | |

19:18 | multiplying four K by one k , you multiply the | |

19:20 | four and the one together . That's four . Multiply | |

19:22 | the K and the K together when multiplying powers at | |

19:25 | the same base to keep the base , add the | |

19:26 | exponents . There's a one on both of them . | |

19:28 | One plus one is two four K times . Negative | |

19:31 | three is negative . 12 K , right ? Positive | |

19:34 | times . Negative is negative Over here . I have | |

19:37 | to do negative two times k squared . That's negative | |

19:40 | . Two k squared . I have to do negative | |

19:42 | two times negative three k negative times negative is positive | |

19:46 | . Six K and I also have to do negative | |

19:48 | two times positive for like , there's me negative eight | |

19:52 | here . If you don't see a number in front | |

19:53 | of the brackets , there's an invisible one there , | |

19:55 | so I have to do negative one times . Case | |

19:57 | where ? Negative . One times negative . Five . | |

19:59 | So that gives me negative one . Okay , squared | |

20:01 | and negative . One times negative . Five is positive | |

20:03 | . Five . Collect your like terms . Do the | |

20:06 | highest degree terms . First to have a four K | |

20:08 | square negative tooth K square . Negative one case square | |

20:11 | . Let's collect those together . Four minus two is | |

20:13 | two minus . One is one . So I have | |

20:15 | one k squared . I'll just write that as K | |

20:18 | squared . Next , I have a negative 12 k | |

20:21 | . A positive six K negative . 12 plus six | |

20:24 | is negative . Six . Alright , negative six K | |

20:27 | and lastly , my constant terms . I've got a | |

20:31 | negative eight plus five negative . Eight plus five is | |

20:34 | negative . Three . So alright , minus three , | |

20:37 | and that's done . I can't collect any of those | |

20:39 | three terms together because they're not like terms . You | |

20:42 | probably have learned this section before distributive property , we | |

20:45 | learned , adding and subtracting polynomial is I think it's | |

20:47 | easier to do this section after , um , you | |

20:51 | know , distributive property . Uh , so basically , | |

20:54 | if you have a set of brackets and you don't | |

20:55 | see a number out front , there's a one there | |

20:58 | . Here , there's a one there . There's a | |

21:00 | one there now to get rid of the brackets . | |

21:02 | You multiply everything in the brackets by the number out | |

21:04 | front . So one times X and one times negative | |

21:06 | six . That's not going to change anything . That's | |

21:08 | just going to give us X minus six . But | |

21:10 | here , if there's a negative one out front multiplying | |

21:14 | both by negative one , it's just going to change | |

21:16 | . The sign of both terms in the brackets . | |

21:18 | Negative one times two is negative . Two negative , | |

21:20 | one times negative . Five acts as positive . Five | |

21:22 | acts . So what happens is both of these terms | |

21:24 | change side . The signs change here . I have | |

21:28 | positive one times X positive one times four . That's | |

21:31 | not gonna change anything . Multiplying things by one doesn't | |

21:33 | change anything . Collect your like terms . I have | |

21:37 | a one x plus five x plus one x , | |

21:40 | that is seven X and now my comments in terms | |

21:44 | of negative six . Take away two plus four negative | |

21:47 | six . Take away to his negative . Eight plus | |

21:49 | four is negative . Four . So I have minus | |

21:51 | four and the expression is fully simplified . Can't collect | |

21:55 | those together because they're not like terms . Okay , | |

21:58 | let's move on to the last thing you would have | |

21:59 | learned in the first unit In grade nine , you | |

22:02 | have learned how to solve equations . And basically solving | |

22:04 | equation means to figure out what value of the variable | |

22:07 | makes the equation true . So here we have X | |

22:09 | Plus four equals seven . That means something plus four | |

22:11 | is equal to seven . You can probably guess the | |

22:13 | answer is three because three plus +47 and that's the | |

22:15 | right answer . But for more complicated questions like when | |

22:18 | we get to questions like here , here , you're | |

22:20 | going to want to be able to solve these algebraic | |

22:24 | lee using probably the balanced method is a method of | |

22:27 | teacher would have taught you first , so basically you're | |

22:29 | allowed to solve this equation . Figure out the value | |

22:31 | box makes it true by isolating the variable by moving | |

22:34 | all of the other numbers away from the X on | |

22:37 | the other side of the equation until you have X | |

22:39 | by itself and then you'll have your answer . So | |

22:41 | to isolate the X , we want to move everything | |

22:44 | away from it . So right now there's a plus | |

22:45 | four on the same side of the equation . With | |

22:47 | it , we can remove that plus four by subtracting | |

22:50 | for as long as we do the same thing to | |

22:53 | both sides of the equation that keeps the equation balanced | |

22:56 | . And we're allowed to do anything we want to | |

22:58 | the equation , as long as we do the same | |

22:59 | thing to both sides have subtracted four from both sides | |

23:02 | , so the equation is still equal . Both sides | |

23:05 | are still equal to each other because I've done the | |

23:06 | same thing to both sides . And look what happens | |

23:08 | if we subtract four from both sides on the left | |

23:11 | , I have four minus four that zero . So | |

23:12 | all that's left on the left side of the equation | |

23:14 | is now . The X on the left side of | |

23:16 | the equation of seven minus four and seven minus four | |

23:19 | is three , so three is the correct answer . | |

23:21 | And don't forget , you can plug this answer back | |

23:23 | into your equation . To check that worked is three | |

23:25 | plus four equal to seven . Yes , you have | |

23:28 | the right answer . Now you should notice that a | |

23:31 | trick to figure out how to isolate the variable . | |

23:33 | Right now , this G five is being subtracted from | |

23:36 | it used to the opposite of subtracting five , which | |

23:39 | is adding five . So if we add five , | |

23:41 | don't forget to do it to both sides of the | |

23:43 | equation . Never do the one side you have to | |

23:44 | do to the other . If I add five to | |

23:47 | both sides of the equation on the left I have | |

23:49 | negative five plus five that zero . So all that's | |

23:51 | left on the left side is G and on the | |

23:53 | right , I have negative three plus five negative . | |

23:57 | Three plus five is too . So that's my final | |

23:59 | answer . G is equal to two and you can | |

24:01 | check your answer by plugging it back into original equation | |

24:03 | . Is two minus five equal to negative three ? | |

24:05 | Yes , So we have the right answer here . | |

24:08 | This is different because it's not five plus you . | |

24:10 | It's five times you . We have five you . | |

24:12 | So to isolate the you that is currently being multiplied | |

24:15 | by five , we do the opposite of multiplying by | |

24:17 | five , which is dividing by five . Remember , | |

24:19 | you have to balance the equation . Whatever you do | |

24:21 | , one side you have to do to the other | |

24:24 | . And then on the left side of the equation | |

24:25 | , we have five divided by five , which is | |

24:28 | one . So those cancel out . So what you | |

24:31 | have left , you have just a you on the | |

24:34 | left side and on the right . We have negative | |

24:36 | 20/5 . And what is negative ? 20 . Divided | |

24:39 | by five . It is negative . Four . Don't | |

24:41 | forget , you can check your answer five times . | |

24:43 | Negative . Four is negative . 20 . So it's | |

24:45 | the right answer . Okay , let's move on to | |

24:47 | to step equations . First thing you want to do | |

24:49 | is you want to isolate the term that has the | |

24:52 | variables . We want to isolate the seven . Why | |

24:54 | ? By moving this positive eight to the other side | |

24:56 | by subtracting eight . So what we're going to do | |

24:59 | Subtract eight from both sides of the equation and on | |

25:02 | the left , we have eight minus 8.0 . It's | |

25:05 | gone . So all we have left on the left | |

25:07 | side equation is seven y on the right . We | |

25:09 | have 15 minus eight and 15 minus eight is seven | |

25:13 | . So now we have seven y equals seven right | |

25:16 | now , otherwise being multiplied by seven . To separate | |

25:18 | a coefficient from a variable like this , we have | |

25:20 | to divide both sides by the seven . And these | |

25:23 | sevens on the left . Cantaloupe because seven divided by | |

25:26 | seven is one . So all we have left is | |

25:27 | y equals 7/7 and 7/7 is one . So there's | |

25:33 | a final answer . You can double check if we | |

25:36 | plug one into this equation . Eight plus seven times | |

25:38 | one is 15 . We have the right answer . | |

25:41 | Okay . What if we have an equation where we | |

25:43 | have more than one term that has a variable ? | |

25:45 | We want to start by getting both of those terms | |

25:47 | to the same side of the equation . Now , | |

25:49 | I want to point something out here for these previous | |

25:52 | questions What you might have noticed you could have done | |

25:55 | instead of using balanced method , you could have thought | |

25:57 | of just moving things to the other side of the | |

26:00 | equation by doing the opposite operation . Like we can | |

26:02 | move this plus four over by making it a minus | |

26:07 | four . Right . You can move something to the | |

26:08 | other side of the equation as long as you apply | |

26:10 | the opposite operation , right ? And that would give | |

26:13 | us what we had . Seven minus four , Axis | |

26:16 | three . So it's a look here . We want | |

26:18 | to get all the terms with the same with the | |

26:20 | variable on the same side of the equation . So | |

26:22 | I want to move , actually going to move the | |

26:25 | other way . I'm going to bring the negative attacks | |

26:27 | to the right side of the equation and get all | |

26:29 | the terms that don't have a variable to the other | |

26:32 | side . So I'm going to bring the plus 15 | |

26:34 | to this side by applying the opposite operations . So | |

26:37 | that just means we're going to change the sign of | |

26:39 | the term . So on the left side of the | |

26:42 | equation , I'm gonna leave the negative five . I'm | |

26:44 | gonna bring this plus 15 over , and it becomes | |

26:47 | a minus 15 on the right side of the equation | |

26:50 | . I'm going to leave the two acts . I'm | |

26:51 | going to bring this negative attacks over which is going | |

26:55 | to change . The sign of the term becomes a | |

26:56 | plus . A duck's now collect my life terms and | |

27:01 | then isolate the variable . Right now , the X | |

27:03 | is being multiplied by 10 the opposite multiplying by 10 | |

27:06 | divide by 10 , so divide both sides by 10 | |

27:09 | to keep it balanced . The tens cancel and what | |

27:11 | I have is negative . 20/10 equals X negative 20 | |

27:15 | divided by 10 . I know that's negative to negative | |

27:18 | . Two is my answer . And you could plug | |

27:20 | that back in and check your answer here . If | |

27:23 | you have brackets , let's start off by getting rid | |

27:26 | of the brackets by using your distributed property that you | |

27:28 | would have learned previously . So start by getting rid | |

27:31 | of the brackets by multiplying the term out front by | |

27:34 | all the terms inside the brackets . Notice I didn't | |

27:36 | distribute this negative because it's not in the brackets , | |

27:38 | so I have four . X plus 12 equals two | |

27:41 | X plus 12 minus eight . Um , now we | |

27:46 | want to I'm gonna simplify this 12 minus eight , | |

27:49 | if you don't mind . Positive 12 . Take away | |

27:52 | eight . That's positive for So I'm just gonna simplify | |

27:55 | that quickly . I'm going to get all the terms | |

27:57 | of the very well on the same side , so | |

27:58 | I'm gonna bring this positive two X to the left | |

28:01 | becomes a minus two X . I'm gonna bring the | |

28:03 | constant terms of the rights of the positive four stays | |

28:06 | on the right . Bring the plus 12 over . | |

28:07 | Becomes a minus 12 . Collect and then isolate the | |

28:12 | variable . The X is being multiplied by two . | |

28:16 | So divide both sides by two . These two is | |

28:18 | canceled because two divided by two is one and what | |

28:21 | I have I have just an X on the left | |

28:24 | on the right . I have negative divided by two | |

28:26 | , which is negative . Four . And you can | |

28:28 | Don't forget . You can double check your answer by | |

28:30 | plugging into the equation here to make sure you have | |

28:32 | the right answer . Okay , Here , let's look | |

28:36 | at fractions . So I have C divided by three | |

28:39 | . Well , don't forget if we want to move | |

28:40 | that divided by three to the other side , what's | |

28:42 | the opposite of dividing by three , while the opposite | |

28:44 | of divided by three is multiplying by three . So | |

28:46 | I'm gonna multiply both sides of the equation because we | |

28:48 | have to keep a balance by three . So I | |

28:51 | rewrote the equation but multiply both sides by three . | |

28:53 | Why did I do that ? Because here I have | |

28:55 | a three divided by three equals one . So they | |

28:58 | cancel out . So all I have left is C | |

29:00 | equals three times two which is six and we can | |

29:03 | check is six divided by three equals two . Yes | |

29:06 | , we have the right answer . What if it | |

29:08 | looks like this ? Whenever you see a fraction an | |

29:10 | equation , you can get rid of the fraction by | |

29:12 | multiplying both sides of the equation by whatever the denominator | |

29:15 | is . So I'm going to multiply the left side | |

29:18 | by four and I'm going to multiply the right side | |

29:22 | by four . Whatever you do , one side you | |

29:24 | have to do to the other . And why did | |

29:26 | I choose four ? Because four divided by four is | |

29:29 | one . So they cancel out . So what I | |

29:31 | have is one times X minus three , which is | |

29:33 | just X minus three on the other side , four | |

29:36 | times negative two , which is negative . Eight . | |

29:38 | Move this minus three to the other side . Or | |

29:40 | think of balance . Method opposite of subtracting three is | |

29:44 | adding three . So add three to both sides on | |

29:47 | the left . Negative three plus 30 So I just | |

29:49 | have X equals negative eight plus three negative eight plus | |

29:53 | three is negative . Five . Let's do another question | |

29:57 | with multiple fractions . So we have multiple fractions here | |

30:01 | . If you have more than one fraction you can | |

30:03 | get rid of both them at the same time . | |

30:05 | If we multiply both equations by a common denominator . | |

30:08 | So what's a common multiple between three and five ? | |

30:11 | So if we were to count by threes and count | |

30:12 | by fives , what's the first number they would have | |

30:14 | in common ? It would be 15 . So what | |

30:16 | we're going to do is multiply both sides by 15 | |

30:21 | . Remember , whatever you do to one side , | |

30:22 | you have to do to the other side to keep | |

30:24 | it balanced . So I multiplied both sides by 15 | |

30:28 | and you'll see how that works . Right here on | |

30:31 | the left have 15 divided by 3 15 , divided | |

30:33 | by three is five on the right . I have | |

30:36 | 15 , divided by 5 15 , divided by five | |

30:39 | is three . So now my fractions are gone . | |

30:41 | So I have five times , one times X plus | |

30:43 | four . So I'll just write five times X plus | |

30:45 | four . And here I have three times one times | |

30:48 | X plus two . So three times one is three | |

30:51 | . So three times X plus two . And now | |

30:53 | you know how to solve it from here . I'll | |

30:55 | just do it very quickly . Get ready your brackets | |

30:57 | using distributed property five x plus 20 equals three X | |

31:02 | plus six . Get all the terms of the variables | |

31:03 | on the same side . I'll bring them to the | |

31:05 | left . I'll bring the three x over , becomes | |

31:07 | a minus three x , Leave the constant of six | |

31:09 | . Bring the plus 20 or becomes a minus 20 | |

31:12 | . And I have two . X equals negative . | |

31:15 | 14 . Oops , minus 20 two X equals negative | |

31:23 | 14 . Mm hmm . And then isolate the X | |

31:28 | by dividing both sides by two . Because that X | |

31:31 | is being multiplied by two right now , make sure | |

31:33 | it stays balanced . Those two's cancel and I have | |

31:36 | X equals negative 14/2 , which is equal to negative | |

31:41 | seven . So there's my final answer . X equals | |

31:44 | negative . Seven . What I have here for this | |

31:47 | next example . I have a special case that we | |

31:50 | can use here . Now . We could multiply both | |

31:52 | sides by 30 to get rid of the brackets . | |

31:54 | That works fine , but I have a shorter way | |

31:56 | . If we have a fraction equals fraction and that's | |

32:01 | it . Nothing else added off at the end here | |

32:04 | anywhere else . If you have fraction equals fraction , | |

32:06 | you can use a shortcut . You can use what's | |

32:10 | called cross multiplication , you can multiply the denominator of | |

32:12 | one side by the numerator of the other . So | |

32:16 | six times X minus four is equal to on the | |

32:20 | other side of the equation , right ? The product | |

32:22 | of the denominator of the other fraction by the numerator | |

32:24 | of the other one . So five times X minus | |

32:27 | three and then sold from here . Distributive property six | |

32:32 | x minus 24 equals five X minus 15 . Get | |

32:38 | all the terms of the variable on one side . | |

32:39 | Six X minus five X equals negative . 15 plus | |

32:43 | 24 we get X equals nine . There's our final | |

32:48 | answer . You could plug back in and check now | |

32:53 | . Cross multiplication is nice . It's a nice shortcut | |

32:55 | , but make sure you only use it when you | |

32:57 | have fraction equals fraction . That's the only time it | |

32:59 | works . This if you can do this question , | |

33:02 | you're good with solving equations . So at this question | |

33:05 | here , what we need to do is get rid | |

33:06 | of all three fractions and on one side of the | |

33:09 | equation , we have multiple terms . So we're gonna | |

33:12 | have to find a common denominator between 23 and four | |

33:15 | . And that would be 12 . So we're gonna | |

33:18 | have to multiply both sides of the equation by 12 | |

33:22 | if we want to get rid of the fractions . | |

33:26 | So I'm going to apply both sides by 12 . | |

33:28 | And what happens over here ? Since I have more | |

33:30 | than one term on this side , this 12 is | |

33:32 | going to get multiplied by both terms . So what's | |

33:36 | going to look like I'm gonna have to do 12 | |

33:39 | times X plus one over to you , plus 12 | |

33:43 | times two X plus 3/3 equals 12 times X over | |

33:50 | four . And then from there we can simplify . | |

33:54 | 12 Divided by 26 12 , divided by three is | |

33:59 | four and 12 , divided by four is three . | |

34:02 | So I have no more fractions . I have six | |

34:04 | times x plus one plus four times two x plus | |

34:10 | three equals three times X , which I'll just raise | |

34:13 | three X and then you're good solving from here . | |

34:16 | Distributive property . Um , eight X plus 12 equals | |

34:22 | three X . I'm gonna clock some like terms on | |

34:24 | the left . Here for six x plus eight x | |

34:26 | 14 x six plus 12 is 18 . I'm going | |

34:31 | to get all the terms of the X on the | |

34:32 | same side . I'm gonna bring the positive three x | |

34:35 | or becomes a negative three x , Bring the plus | |

34:37 | 18 or becomes a negative 18 . So have 11 | |

34:39 | x equals negative 18 and what I have here , | |

34:43 | two by both sides by 11 to get the X | |

34:46 | by itself . And what I have is X equals | |

34:52 | negative 18 over 11 , and that's a fine answer | |

34:56 | . A fraction is a perfectly fine answer . Just | |

34:58 | make sure it can't be reduced any further . X | |

35:00 | is negative . 18/11 . Okay , last thing probably | |

35:06 | you would have done in the solving equations is look | |

35:08 | at some word problems . I'm going to do a | |

35:10 | very quick one just to remind you how to do | |

35:11 | it . So two friends are collecting pop can tabs | |

35:14 | , and Tasha has 250 more than Christian . Together | |

35:16 | , they have 880 pop can tabs . How many | |

35:18 | pop can tabs to each friend has each friend collected | |

35:21 | ? So the question wants to know . How many | |

35:23 | pop can tabs does Natasha have ? How many pop | |

35:27 | can tabs does Kristen have ? So that's the question | |

35:30 | is asking us so we want to start by making | |

35:32 | a polynomial expression for each of those people . Natasha | |

35:35 | has 250 more than Christian . We don't know how | |

35:38 | many Christian has , so we use a variable for | |

35:40 | Kristen and we know Natasha has 250 more than X | |

35:43 | . So x plus 2 50 represents that scenario . | |

35:45 | And then we write an equation using these two expressions | |

35:50 | . Together they have 880 . So I know if | |

35:52 | I add those two expressions together , So if I | |

35:55 | do X plus X plus 2 50 I know that | |

35:59 | that should equal 880 . Now I can solve this | |

36:04 | equation one X plus one access to X . Bring | |

36:07 | that to 50 to the other . Side becomes a | |

36:09 | minus 2 50 . So what I have is two | |

36:12 | x equals 6 30 then divide both sides by two | |

36:17 | . To get rid of that coefficient of two . | |

36:19 | I have X equals 3 15 . So what does | |

36:23 | that tell us ? It tells us Kristen has 315 | |

36:27 | and Natasha members to 50 more than X , and | |

36:30 | no X is 3 15 . So what I have | |

36:32 | is 5 65 and I know together those two numbers | |

36:36 | add to 80 . So I have the correct answer | |

36:38 | here . Okay , That's it for algebra . Um | |

36:41 | , watch the next two parts Watch part two Linear | |

36:43 | relations in part three . Geometry geometry will be the | |

36:46 | shortest section . This is the longest section of grade | |

36:49 | nine . So make sure you watch the next two | |

36:51 | sections and you've learned all of grade nine in under | |

36:53 | one hour . |

#### DESCRIPTION:

Here is a great exam review video reviewing all of the main concepts you would have learned in the MPM1D grade 9 academic math course. The video is divided in to 3 parts. This is part 1: Algebra. The main topics in this section are exponent laws, polynomials, distributive property, and solving first degree equations. Please watch part 2 and 3 for a review of linear relations and geometry. If you watch all 3 parts, you will have reviewed all of grade 9 math in 60 minutes. Enjoy! Visit jensenmath.ca for more videos and course materials.

#### OVERVIEW:

ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 1) is a free educational video by Lumos Learning.

This page not only allows students and teachers view ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 1) videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.

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