ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 1) - Free Educational videos for Students in K-12 | Lumos Learning

ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 1) - Free Educational videos for Students in k-12


ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 1) - By Lumos Learning



Transcript
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 .
Summarizer

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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.

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ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 1) is a free educational video by Lumos Learning.

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