ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 1)

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okay here's a video that's going to

review all main concepts you learned in

grade 9 math in under 60 minutes you're

going to learn all grade 9 in under an

hour this is a great video if you're

doing exam review or if you're about to

take the course and you just want to

know what you're going to learn now in

this video I'm not going to go in-depth

into any of the topics I'm going to go

through a couple examples of each topic

but if you want in-depth explanations

and proofs of each topic go to Jensen

math GA and there's video lessons for

each topic there so I'm going to divide

this video into three parts based on the

three main units you do in grade 9 so

the first part is going to be on algebra

and algebra you learn about polynomials

you learn how to collect like terms you

learn distributive property learn

exponent laws and the next part is

solving equations so I'm going to go

through that part first and watch the

second video if you want to learn about

linear relations and a third video if

you want to review the geometry section

so watch all three videos and you'll

have reviewed all of the main concepts

of grade 9 math in under an hour so

let's get started with the first topic

right away I'm going to go quickly just

so I can get this done under an hour

so the first main topic you would have

learned in grade 9 math is exponent laws

so to do exponent laws first main thing

you have to remember you have to know

what a power is so let's say we have

five squared we have to know that that

means five times five right this is the

base five is the base 2 is the exponent

and this is a power so the exponent

tells you how many of the base so how

many factors of the base are being

multiplied together so there are some

exponent laws you would have learned the

product of powers rule so if you have

two powers being multiplied together the

dot means multiply so two powers being

multiplied together have the same base

that's important the base has to be the

same there's a rule you keep the base

the same and you add the exponents so

that equals x to the a plus B so you can

rewrite as a single power by keeping the

base the same adding the exponents what

if you have two powers the same base

being divided by each other there's a

similar rule for that one but similar

rule you keep the base the same and this

time you subtract the exponents that's X

to the a minus B you would have learned

the power of the power rule so if you

have it exploring on top of an exponent

can keep the base and you multiply the

exponents X to the a times B what if you

have a power where the base is the

quotient so a power of a quotient

you have to remember this exponent gets

put on to the numerator and the

denominator so this is equal to a to the

x over B to the power of X what if we

have a power where the base is a product

of numbers or variables so this exponent

outside of the brackets gets put on to

each of the factors inside the brackets

so this would be equal to a to the x

times B to the X and a couple more rules

you would have learned that anything to

the power of zero is equal to one and

the negative exponent rule if you have

power with a negative exponent you can't

leave a final answer with a negative

exponent in this section so what you

have to do is rewrite this the positive

exponent by writing the reciprocal of it

so it would be 1 over X to the power of

positive a so I'm going to go through a

couple examples that review all these

rules that these are the main exponent

rules you would have learned in grade 9

so let's practice a few of them without

variables first and I'll throw some

variables into the mix so here we have

powers that all have the same base right

y'all the base of for the first two are

being multiplied together so what I can

do is rewrite as a single power by

adding the exponents 3 plus 5 is 8 and

then I have to divide that by 4 squared

so I can rewrite that since they have

the same base and they're being divided

by keep the base the same and I subtract

the exponents so I end up at 4 to the

power of 6 and then after I've written

it as a single power I can evaluate it

for the power of 6 is 4096 number for

the power of 6 means 4 times 4 times 4

times 4 times 4 times 4

it doesn't mean 4 times 6 okay power of

a quotient and we have power of a power

here as well so let's start off with a

power to power rule so I have an

exponent on top of an exponent I can

rewrite this by multiplying these powers

together 2 times 3 is 6 and notice I put

could have put it into the denominator

if I wanted to you but don't put it into

both I decided to put it to the

numerator to make it easier for me here

so here I have negative 2 over 3 to the

power of 6 so what I have now is power

of a quotient so remember 4 power of a

quotient the exponent if the base is a

quotient the exponent of the power goes

onto the numerator and the denominator

so this equals negative 2 to the 6 over

3 to the 6

and then I have to let me move this down

and then I have to evaluate each of

those powers so negative two to the

power of six is 64 and three to the

power of 6 is 729 so that's fully

simplified I can't reduce that fraction

any further let's look at these two here

so I have product of powers here and the

powers have the same base so I can use

my exponent want to simplify by keeping

the same base and adding the exponents 4

plus negative 7 be careful with your

integers 4 plus negative 7 is negative 3

but I can't leave that as my final

answer with the negative exponent what I

have to do is rewrite that with a

positive exponent by writing the

reciprocal of it so 1 over 5 to the

power of positive 3 and now I can

evaluate 5 to the power of 3 and that

gives me 125 so 1 over 125 is my final

quotient of powers so what I have to do

I can rewrite it as a single power by

keeping the base and subtracting the

exponents 7 minus 7 0 and remember

anything to the exponent of 0 is equal

to 1 and you should notice up here when

you have something divided by itself

that answer is always going to be one as

well so think of it either way let's

throw some variables into the mix so I

have a product of powers here I have

these three powers being multiplied

together they're all the same base so I

can rewrite as a single power by keeping

the same base and adding the exponents 3

plus 4 plus 5 that's 12a to the 12

I can't evaluate it evaluate that since

it is a variable sweet leave it like

that there's our final answer what we

have here is power of a product so the

base of the power is a product of 4x

squared and Y to the 5 so what I have to

do is put this outside exponent onto

each of the factors in the product onto

the 4 onto the x squared and on Y to the

fly so I have to do 4 cubed I'll put in

brackets just so it looks nicer here I

have to do x squared cubed and I have to

do Y to the 5 cubed write this outside

exponent goes on to each of the factors

of the product of the base and now I can

evaluate 4 cubed is 64 X to the 2 to the

3 well that's power the power I don't

have to multiply the exponents there so

that's X

to the six and Y to the five to the

three power of a power rule again Y to

the 15 and that's done here I have

product of some powers and of some

coefficients here so let's start off

with the five times four always start

with your coefficients we can multiply

five times four just as you always would

it's 20

now you look for powers that have the

same base I have an M to the five and an

M Squared those have the same base so I

know when I multiply powers to add the

same base I keep the base the same and I

add the exponents five plus two is seven

so I've m2 the seven also I have two

powers of M that are being multiplied so

I can simplify that by writing as a

single power by adding the exponents and

keep in mind this M has an exponent of

one even though you don't see an

exponent there so one plus four is five

and that is fully simplified there that

question is done let's look at a

quotient of powers here same thing stare

divided by 27 and divide 36 by 27

exactly as you always would we don't

want a decimal answer we just want to

reduce it right 27 doesn't go into 36

evenly but I can reduce that fraction by

finding number that goes into both of

them evenly and in fact nine goes into

both 36 and 27 9 goes into 36 four times

and it goes into 27 three times so I can

reduce 36 over 27 to four over three now

I look at my variables so I have some

powers but with the same bases I have an

X cubed and X to the six that are being

divided by each other so I know I can

simplify that writing as a single power

if I subtract the exponents 3 minus 6 is

negative 3 and make sure you put your

quotients into the numerator here we'll

take care of that negative in a second

next we also have two powers of a base

of why Y to the 9 divided by y to the 4

rewrite it's a single power by keeping

the base and subtracting the exponents 9

minus 4 is 5 and remember always be your

quotients in to the numerator

now we have to do something with this

power that has a negative exponent this

X to the negative 3 is a power with a

negative exponent we can take care of

that negative by writing the reciprocal

of it and what's going to happen is this

X to the negative 3

not before just the X to the negative 3

just that power if we

bring it into the denominator it'll make

the exponent positive so what we have

for our final answer the 4 and the y to

the 5 stay in the numerator the 3 stays

in the denominator and we bring the

power of X to negative 3 to the

denominator and it makes it a power of

positive 3 so there's my final answer

there let's do one more if you can do

this example you can be confident that

you understand all of the exponent laws

therefore now let's look at just the

numerator and simplify that I have an 8

times a 4 that's 32 I have now look for

powers the same base of a b cubed times

a B to the 1 keep the base add the

exponents remember product rule when

multiplying powers to in base you add

the exponents and now I have a d to the

1 times a D to the two that's D to the 3

if we look in the denominator let's just

leave this two out front for now and

let's just do this power of a product

rule here where this exponent of 2 has

to go on each factor of the base has two

on the to the B and the D so we have to

square the two which makes it a 4 we

have to square the B which makes it a B

squared and square the D which makes it

a d squared so now I have 2 times 4 B

squared d squared so I can simplify that

2 times 4 and make it an 8 so be the 4

leave the numerator for now 2 times 4 8

and then at b squared d squared and last

step let's do or dividing 38 32 divided

by 8 divide those just like you would

divide any old numbers 32 divided by 8

goes into 32 4 times so my answer for

now do B to the 4 divided by B to the 2

when dividing powers the same days keep

the base subtract the exponents D to the

3 divided by D to the 2 once again keep

the base subtract the exponents it's a d

to the 1 if it's a 1 you don't have to

write the 1 alright that's it for

exponent law review let's move on to

polynomials quickly so basically you

have to remember what a term is a term

is an expression form of a product of

numbers and variables so for example 2x

that's a term it's a product of a number

2 and a variable X and what a polynomial

is such an expression

interesting on one or more of these

terms that are connected by addition or

subtraction operators so for example I

could have the polynomial 4x squared

plus 3x plus 1 that's a polynomial where

we have three terms term 1 2 3 they're

signs now we can classify polynomial

based on how many terms as by name if

the polynomial only has one term like

this 2x up here we call that a monomial

if a polynomial has two terms we call it

a binomial if it has three terms we call

it a trinomial and if it has anything

more than that there's no special name

for it like that's four terms we just

simply call it a four term polynomial if

it had five we would call it a five term

polynomial and so on so one more thing

you'll have to be able to do is state

the degree of a term and the degree of a

polynomial now just state the degree of

a term by looking at one individual term

you can state the degree of it just by

adding up so finding the sum of the

exponents on all the variables in that

term so if we look at this first example

here three x squared Y this is one term

so this is a monomial there's nothing

else added or subtracted from it it's

one term has two variables so to find

the degree of this term we add the

exponents on the variables Q minus y has

a one so on the X and on the Y we add

the exponents 2 plus 1 is 3 so this this

right here this term right here is

degree 3 if we look at this next example

here what we have here is three

different terms being added together so

we call this a trinomial we could find

the degree of each term like this first

term here is degree 5 this first term

here is degree 4 because that has an

explorative one and this third term is

degree 6 so if we want to figure out the

degree of the entire polynomial all we

have to do is figure out the degree of

the highest degree term in this

polynomial we don't add all these

degrees together we just pick the term

that has the highest degree so in this

case it would be the third term its

degree 6 so what we say is the entire

polynomial is degree 6 the degree of the

polynomial is

equal to degree of the highest degree

term in the polynomial here we have a

binomial right two terms one to subtract

separated by a subtraction sign the

first term is degree seven right one

plus five plus one is seven

this term is degree six so to find the

degree of the entire polynomial it's

equal to the degree of the highest

degree term this term is the higher

degree so the degree of the entire

polynomial is seven let's look at

collecting like terms now so first of

all what are like terms like terms are

terms of the exact same variables with

the exact same exponents for example

these two terms they have the exact same

variables they both have an X and a Y

and on those variables are the exact

same exponents on the X's of 2 and on

the Y is a 1 on both of them so those

are like terms the fact that the

coefficients are different do not make

them not like terms the coefficients

don't matter when deciding if they're

like terms or not you just look at the

variables and the exponents so if we

look at these two here these are not

like terms

why because this one has a Y to the

power of one this one has a Y to the

power of two so they don't have the same

variable so the same exponents so

they're not like terms why do we have to

be able to know what like terms are

because we can we can collect like terms

together for example here I have four

terms so when better being added or

subtract from each other when adding or

subtracting terms we can collect them

together by adding or subtracting the

coefficients only and keeping the

variable the same right this is

different than exponent laws with

exponent laws we were using them when we

are multiplying powers or dividing

powers but now we're going to be adding

or subtracting terms and we can collect

them we can collect like terms together

so what we want to first do is group the

terms they're like terms together and

make sure the sign that is to the left

of the term stays with it so for example

I have a negative 2x and a negative 5x

those are like terms so let's write

those beside each other keeping the

signs that are to the left of them and I

also have a positive seven line a

negative 9y let's write those beside

each other because those are like terms

now I can collect them together I can

collect the negative 2x minus 5x

together because they're like terms by

tracking the coefficients only so

negative 2 minus 5 thats negative 7 and

then you keep the variable exactly the

same don't change the exponent on the

variable at all it stays exactly the

same don't get this confused with

exponent laws now look at this group of

like terms the 7y minus 9y 7 minus 9 is

negative 2 so I write negative 2y so

that is fully complete I can't collect

these two terms together because they're

not like terms so that expression is

fully simplified let's look at the next

expression here it's a little bit longer

find the groups of like terms I have a 3

x squared Y and an 8 x squared Y those

both have the exact same variables with

the same exponent so I'm going to write

those beside each other 3x squared y 8 x

squared Y I have a 4y and a negative 1y

so I'll write my positive 4y a my

negative 1y remember the coefficient is

1 if you don't see it and I also have

two constant terms constant term means

the term without a variable those are

like terms I have a positive 7 and a

positive 80 right those beside each

other now collect your like terms so 3x

squared y plus 8x squared Y just add the

coefficients only 11x squared Y don't

change the exponents on the variables at

all when you are adding or subtracting

like terms together I have a positive 4y

minus 1y that's positive 3y and I have

positive 7 plus 80 that's positive 87

that expression is fully simplified none

of these three things are like terms

with each other so I can't collect them

together don't try and go any further

than you can and notice the order I

wrote these in you should do it in this

order the highest degree term goes first

and then it goes in descending order

this term is degree 3 right 2 plus 1 is

3 this term is degree 1 and a constant

term is degree 0 so the degrees should

go in descending order okay it's like a

distributive property so basically for

multiplying a monomial by a polynomial

in this case a binomial this is how you

do it everything inside the brackets

gets multiplied by the term out front of

the brackets and that gets rid of the

bracket so if I want to do a times X

plus y I have to do a times X right here

and I have to do a times y here so for

example if I 5 times 4x plus 2 I need to

multiply the 4x

and the to both of them by the five

that's out front

so five times four X is 20 X 5 times 2

is positive 10 there's my solution those

can't be collected together because

they're not like terms so let's practice

distributive property so here I'm doing

a monomial multiplied by a trinomial so

everything in the brackets needs to be

multiplied by the term out front be

careful with your signs and then that

will get rid of the bracket so I have to

do negative 3 times 2x squared well

negative 3 times 2 is negative 6 so I

have negative 6x squared I have to do

negative 3 times negative 5 X don't

forget this sign belongs to this 5x so

negative times negative is positive 15 X

and after your negative 3 times positive

4 that's negative 12 and I can't collect

any of these 3 terms together because

they're not like terms how about here

I'm going to have to multiply the X plus

3 by the 3 out front I'm going to have

to multiply this X and this one by the 2

up front that will get rid of the

brackets so 3 times X is 3x 3 times 3 is

positive 9 2 times X positive 2x 2 times

1 positive 2 now I can collect like

terms here because I have a 3x and a 2x

put those together that's 5x and I have

a 9 and the to put those together that's

positive 11

let's do one more example here for

collecting like terms I have 4 K times K

and times negative 3 now keep in mind

this K inside the brackets think of that

as a 1 K if you want so when multiplying

4 K by 1 K you multiply the 4 and the 1

together that's 4 multiply the K and the

K together with multiplying powers the

same base you keep the base add the

exponents there's a 1 on both of them 1

plus 1 is 2 4 K times negative 3 is

negative 12 K right positive times

negative is negative over here I have 2

negative 2 times K squared that's

negative 2 K squared I have 2 negative 2

times negative 3 K negative times

negative is positive 6 K and I also have

to do negative 2 times positive 4 that

gives me negative 8 here if you don't

see a number in front of the brackets

there's an invisible one there so I have

to do negative 1 times K squared

negative 1 times negative 5 so that

gives me negative 1 K squared and

negative 1 times negative 5 is

positive five collect your like terms to

the highest degree terms first I have a

4 K squared negative 2 K squared

negative 1 K square let's collect those

together 4 minus 2 is 2 minus 1 is 1

so I have 1 K squared I'll just write

that as K squared next I have a negative

12 K a positive 6 K negative 12 plus 6

is negative 6 all right negative 6 okay

and lastly my constant terms I've got a

negative 8 plus 5 negative 8 plus 5 is

negative 3 so I'll write minus 3 and

that's done I can't collect any of those

three terms together because they're not

like terms you probably have learned

this section before distributive

property we learn adding and subtracting

polynomials I think it's easier to do

this section after you know distributive

property so basically if you have a set

of brackets and you don't see a number

up front there's a 1 there here there's

a 1 there there's a 1 there now to get

rid of the brackets you multiply

everything in the brackets by the number

out front so 1 times X and 1 times

negative 6 that's not going to change

anything that's just going to give us X

minus 6 but here if there's a negative 1

out front multiplying both by negative 1

it's just going to change the sign of

both terms in the brackets negative 1

times 2 is negative 2 negative 1 times

negative 5x is positive 5x so what

happens is both of these terms change

side the signs change here I have

positive 1 times X positive 1 times 4

that's not going to change anything

multiplying things by 1 doesn't change

anything

collect your like terms I have a 1 X

plus 5x plus 1 x that is 7 X and now my

constant terms have negative 6 take away

2 plus 4 negative 6 take away 2 is

negative 8 plus 4 is negative 4 so I

have minus 4 and that expression is

fully simplified can't collect those

together because they're not like terms

ok let's move on to the last thing you

would have learned in the first unit in

grade 9 you have learned how to solve

equations and basically a solving

equation means to figure out what value

of the variable makes the equation true

so here we have X plus 4 equals 7 that

means something plus 4 is equal to 7 now

you probably guess the answer is 3

because 3 plus 4 is 7 and that's the

but for more complicated questions like

when we get to questions like here here

you're going to want to be able to solve

these algebraically using probably the

balance method as a method your teacher

would have taught you first so basically

you're allowed to solve this equation

figure out the value of X makes it true

by isolating the variable by moving all

of the other numbers away from the X

onto the other side of the equation

until you have X by itself and then

the X we want to move everything away

from it so right now there's a plus 4 on

the same side of the equation with it we

can remove that plus 4 by subtracting 4

as long as we do the same thing to both

sides of the equation that keeps the

equation balanced and we're allowed to

do anything we want to the equation so

long as we do the same thing to both

sides I've subtracted 4 from both sides

so the equation is still equal both

sides are still equal to each other

because I've done the same thing to both

sides and look what happens if we

subtract 4 from both sides on the left I

have 4 minus 4 that's 0 so all that's

left on the left side of the equation is

now the X on the left side of the

equation I have 7 minus 4 and 7 minus 4

is 3 so 3 is the correct answer and

don't forget you can plug this answer

back into your equation to check your

work is 3 plus 4 equal to 7 yes you have

the right answer now you should notice

that a trick to figure out how to

isolate the variable right now this g5

is being subtracted from it used to the

opposite of subtracting 5 which is

do it to both sides of the equation

whatever you do to one side you have to

do to the other if I add 5 to both sides

of the equation on the Left I've

negative 5 plus 5 that's 0 so all that's

left on the left side is G and on the

right I have negative 3 plus 5 negative

3 plus 5 is 2 so that's my final answer

G is equal to 2 and you can check your

answer by plugging it back into your

equation is 2 minus 5 equal to negative

3 yes so we have the right answer here

this is different because it's not 5

plus u it's five times U we have five

use so it's isolate the U that is

currently being multiplied by five we do

the opposite of multiplying by five

which is dividing by five and remember

you have to balance the equation

whenever you do one side you have to do

to the other and then on the left side

of the equation we have five divided by

five which is one so those cancel out

so what you have left you have just a

you on the left side and on the right we

have negative 20 over five and what is

negative 20 divided by five it is

negative four don't forget you can check

negative 20 so it's the right answer

okay let's move on to two step equations

first thing you want to do is you want

to isolate the term that has the

variable so we want to isolate the 7y by

moving this positive eight to the other

side by subtracting eight so what we're

going to do subtract 8 from both sides

of the equation and on the Left we have

8 minus 8 that's zero it's gone so all

we have left on the left side equation

is 7y on the right we have 15 minus 8

and 15 minus 8 is 7 so now we have 7y

equals 7 right now the y is being

multiplied by 7 to separate a

coefficient from a variable like this we

have to divide both sides by the 7 and

these 7s on the Left cancel out because

7 divided by 7 is 1 so all we have left

is y equals 7 over 7 and 7 over 7 is 1

so there's our final answer you can

double check if we plug 1 into this

equation 8 plus 7 times 1 is 15 we have

the right answer okay what if we have an

equation where we have more than one

term that has a variable what you want

to start by getting both of those terms

the same side of the equation now I want

to point something out here for these

previous questions what you might have

noticed you could have done instead of

using balance method you could have

thought of just moving things to the

other side of the equation by doing the

opposite operation like we can move this

plus 4 over by making it a minus 4 right

you can move something to the other side

of the equation as long as you apply the

opposite operation right and that would

give us what we had 7 minus 4x is 3 so

let's look here we want to get all the

terms with the same width with a

variable in the same side of the

equation so I want to move actually I'm

going to move the the other way I'm

going to bring the negative 8x to the

right side of the equation and get all

the terms that don't have a variable to

the other side so I'm going to bring the

plus 15 to this side by applying the

opposite operation so that just means

we're going to change the sign of the

term so on the left side equation I'm

going to leave the negative 5 and

bring this +15 over and it becomes a

minus 15 on the right side of the

equation I'm going to leave the 2x I'm

going to bring this negative 8x over

which is going to change the sign of the

term becomes a plus 8x now collect my

like terms and then isolate the variable

right now the X is being multiplied by

10 the opposite of multiplying by 10

divided by 10 so divide both sides by 10

to keep it balanced the tens cancel and

what I have is negative 20 over 10

equals X negative 20 divided by 10 I

know that's negative 2 so negative 2 is

my answer and you could plug that back

have brackets let's start off by getting

rid of the brackets by using your

distributive property that you would

have learned previously so start by

getting rid of the brackets by

multiplying the term out front by all

the terms inside the brackets notice I

didn't distribute to this negative

because it's not in the brackets so I

have 4x plus 12 equals 2x plus 12 minus

8 now we want to I'm going to simplify

this 12 minus 8 if you don't mind

positive 12 take away 8 that's positive

4 so I'm just going to simplify that

quickly I'm going to get all the terms

with the variable on the same side so

I'm going to bring this positive 2x to

the left becomes a minus 2x I'm going to

bring the constant terms to the right so

the positive 4 stays on the right bring

the plus 12 over becomes a minus 12

collect and then isolate the variable

the X is being multiplied by Q so divide

both sides by 2 these twos cancel

because 2 divided by 2 is 1 and what I

have I have just an X on the left on the

right I've negative 8 divided by 2 which

is negative 4 and you can don't forget

you could double chance check your

answer by plugging into the equation

here to make sure you have the right

answer okay here let's look at fractions

so I have C divided by 3 well don't

forget if we want to move that divided

by 3 to the other side what's the

opposite of dividing by 3 well the

altitude divided by 3 is multiplying by

3 so I'm going to multiply both sides of

the equation because we have to keep it

balanced by 3 so I rewrote the equation

but multiply both sides by 3 why did I

do that because here I have a 3 divided

by a 3

most ones they cancel out so all I have

left is C equals three times 2 which is

6 and we can check is 6 divided by 3

equal to 2 yes we have the right answer

what if it looks like this whenever you

see a fraction in an equation you can

get rid of the fraction by multiplying

both sides of the equation by whatever

the denominator is so I'm going to

multiply the left side by 4 and I'm

going to multiply the right side by 4

whatever you do one side you have to do

to the other

and why did I choose 4 because 4 divided

by 4 is 1 so they cancel out so what I

have is 1 times X minus 3 which is just

X minus 3 on the other side 4 times

negative 2 which is negative 8

move this minus 3 to the other side or

think of balance method opposite of

both sides on the Left negative 3 plus 3

is zero so I just have x equals negative

8 plus 3 negative 8 plus 3 is negative 5

let's do another question with multiple

fractions so we have multiple fractions

here if you have more than one fraction

you can get rid of both of them at the

same time if we multiply both equations

by a common denominator so what's a

common multiple between 3 & 5 so if we

were to count by threes and count by 5's

what's the first number they would have

in common it would be 15 so what we're

going to do is multiply both sides by 15

remember whatever you do to one side you

have to do to the other side to keep it

balanced so I multiplied both sides by

15 and you'll see how that works right

here on the left I have 15 divided by 3

15 divided by 3 is 5 on the right I have

15 divided by 5 15 divided by 5 is 3 so

now my fractions are gone so I have 5

times 1 times X plus 4 so I'll just

write 5 times X plus 4 and here I have 3

times 1 times X plus 2 so 3 times 1 is 3

so 3 times X plus 2 and now you know how

to solve it from here I'll just do it

very quickly get rid of your brackets

using distributive property 5 X plus 20

equals 3x plus 6 get all the terms with

the variable on one on the same side

I'll bring them to the left so I'll

bring the 3x over becomes 2 minus 3x

leave the constant of 6 bring the plus

2000

comes to minus 20 I have 2x equals

negative 14 whoops and minus 22 x equals

negative 14 and then isolate the X by

dividing both sides by 2 because that X

is being multiplied by 2 right now make

sure it stays balanced those twos cancel

and I have x equals negative 14 over 2

which is equal to negative 7 so there's

my final answer x equals negative 7 what

I have here for this next example I have

a special case that we can use here now

we could multiply both sides by 30 to

get rid of the brackets that works fine

but I have a shorter way if we have

fraction equals fraction and that's it

nothing else add it off at the end here

anywhere else if you have fraction

equals fraction you can use a shortcut

you can use what's called cross

multiplication you can multiply the

denominator of one side by the numerator

of the other so 6 times X minus 4 is

equal to on the other side of the

equation write the product of the

denominator of the other fraction by the

numerator of the other one so 5 times X

minus 3 and then solve from here

distributive property 6x minus 24 equals

5x minus 15 get all the terms of the

variable on one side 6x minus 5x equals

negative 15 plus 24 and we get x equals

9 there's our final answer you could

plug back in and check now cross

multiplication is nice it's a nice

shortcut but make sure you only use it

when you have fraction equals fraction

that's the only time it works this if

you can do this question you're good by

solving equation so at this question

here what we need to do is get rid of

all three fractions and on one side of

the equation we have multiple terms so

we're going to have to find a common

denominator between 2 3 & 4 and that

would be 12 so we're going to have to

multiply both sides of the equation by

12 if we want to

get rid of the fractions so I'm going to

multiply both sides by 12 and what

happens over here since I have more than

one term on this side this 12 is going

to get multiplied by both terms so

what's going to look like I'm going to

do 12 times X plus 1 over 2 plus 12

times 2x plus 3 over 3 equals 12 times x

over 4 and then from there we can

simplify 12 divided by 2 is 6 12 divided

by 3 is 4 and 12 divided by 4 is 3 so I

have no more fractions I have 6 times X

plus 1 plus 4 times 2x plus 3 equals 3

times X which I'll this ray is 3x and

then you're good solving from here

distributive property it X plus 12

equals 3x I'm going to collect some like

terms on the left here for 6x plus 8x is

14 X 6 plus 12 is 18 I'm going to get

all the terms of the X on the same sides

I'm going to bring the positive 3x over

becomes a negative 3x bring the plus 18

over becomes a negative 18 so I have 11x

equals negative 18 and what I have here

divide both sides by 11 to get the X by

itself and what I have is x equals

negative 18 over 11 and that's a fine

answer a fraction is a perfectly fine

answer just make sure it can't be

reduced any further X is negative 18

over 11 okay last thing probably you

would have done in the solving equations

is look at some word problems I'm going

to do a very quick one just to remind

you how to do it so to friends or

collecting Popkin tabs and Tasha has 250

more than Kristen together they have 880

Popkin tabs how many Popkin tabs to each

friend has each friend collected so the

question wants to know how many pop-can

tabs

does Natasha have how many pop-can tabs

does Kristen have so that's what the

question is asking us so we want to

start by making a polynomial expression

for each of those people Natasha is 250

more than Kristen

we don't know how many Kristen has so we

use a variable for Kristen and we noted

Tasha has 250 more than X so X plus 250

represents that scenario and then we

write an equation using these two

expressions together they have 880 so I

know if I add those two expressions

together so if I do X plus X plus 250 I

know that that should equal 880 now I

can solve this equation 1 X plus 1 X is

2x bring that 250 to the other side

becomes a minus 250 so what I have is 2

x equals 630 and then divide both sides

by 2 to get rid of that coefficient of 2

I have x equals 315 so what does that

tell us it tells us kristen has 315 and

Natasha member is 250 more than X and we

know X is 315 so what I have is 565 and

I know together those two numbers add to

a so I have the correct answer here okay

that's it for algebra watch the next two

parts watch part 2 linear relations in

part 3 geometry geometry will be the

shortest section this is the longest

section of grade 9 so make sure you

watch the next two sections and you'll

have learned all of grade 9 in under one

hour