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 .

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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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