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in this video we're going to focus on

solving two-step equations so let's

start with the basics we're going to

cover equations with fractions

parentheses variables on both sides and

even decimals but let's start with the

basic two-step equations so how would

you solve this equation let's say that

3x plus 5 is equal to 17 what's the

first thing that we need to do in this

problem in order to solve this equation

we need to isolate the X variable so the

first thing we need to move is the 5 we

want to get X by itself on the left side

so we need to get rid of the 5 since the

5 is added to 3x let's perform the

opposite operation let's subtract both

sides by 5 5 and negative 5 those two

will cancel so what we have left over on

the left side of the equal sign is 3x 17

minus 5 is 12 now all we need to do is

separate the 3 from the X the opposite

of multiplication is division so we need

to divide both sides by 3 so X is 12

divided by 3 which is 4 and this is the

answer here's another example that you

could try 4x plus 3 is equal to 19 so

based on the last example take a moment

and work on this problem feel free to

pause the video now just like before

we're going to start with subtraction

let's subtract both sides by 3 just to

get rid of the 3 on the left side 3 and

negative 3 they add up to 0 so we just

have 4x on the left side and on the

right side it's going to be 19 minus 3

which is 16 next to separate the 4 from

the X we need to divide both sides by 4

4 divided by 4 is 1 1 X and X you can

just write as X on the right side is 16

divided by 4 which is 4 and so that's

the answer for this problem X is equal

to 4

let's try a similar example but slightly

different so let's say that 17 minus 5x

is equal to 2 if that's the case what is

the value of X so what do you think we

need to do in this problem well we need

to get rid of the 17 first on the Left

we have positive 17 so we need to begin

by subtracting both sides by 17 so these

two will add to 0 and then we could

bring down the negative 5x positive 2

minus 17 that's equal to negative 15

next let's divide both sides by negative

5 negative 5 divided by itself is 1

leaving behind 1x negative 15 divided by

negative 5 is positive doing whenever

you divide by two negative numbers

you're going to get a positive result

and so that is the answer X is equal to

3 now let's move on to our next example

consider this equation 9 is equal to 3

plus X divided by 4 so we have a

fraction what do you think we need to do

and it's problem 1 this problem we could

start by subtracting both sides by 3

just as we've been doing 9 minus 3 is 6

now let's bring down the fraction so 6

is equal to X divided by 4 the opposite

of division is multiplication so since X

is divided by 4 to get rid of that let's

multiply both sides by 4 on the right

side the 4s will cancel leaving behind

an X variable on the Left we have 4

times 6 which is 24 and so that is the

answer now if you want to check your

work you can take this answer and plug

it into the original equation so 9 is

equal to 3 plus let's replace X with 24

24 divided by 4 is 6

and 3 plus 6 is 9 so 9 equals 9 the

equation is balanced which means that

this here is indeed of our answer X is

indeed 24 now let's move on to the next

problem go ahead and take a minute and

try this example so 8 plus X divided by

3 is equal to 12 so pause the video and

write this equation down go ahead and

solve it so I'm going to start off by

subtracting both sides by 8

12 minus 8 is 4 now then the next step

we're going to multiply both sides by 3

so on the left the 3s will cancel on the

right we have 4 times 3 which is 12 and

so that's going to be the answer for

this problem

X is equal to 12 is another one that you

could try so in this example the X

variable is on the right side so go

ahead and try so this problem is very

similar to the first two problems that

we covered even though the X variables

on the other side the same rules apply

let's begin by subtracting both sides by

2 so 14 minus 2 that's 12 and on the

right side these will cancel just as

before and as you can see solving these

equations they're not that difficult now

we'll need to divide both sides by 3 so

X is going to be 12 divided by 3 so X is

4 in this example now sometimes you may

have multiple X variables like in its

problem 3x + 8 + 5 X let's say this is

all equal to 32 now what should we do in

this example now this problem might

require more than two steps so it might

be a multi-step problem which I'll have

more later in this video but what do you

think we need to do in this problem if

you see two X variables on the same side

the first thing you should do is combine

like terms so we should add 3x + 5 X

which is 8 X now we have a problem

that's similar to the ones that we dealt

with in the beginning so let's go ahead

and subtract both sides by 8 8 minus 8

is 0 we could bring down the 8x and on

the left side we have 32 minus 8 which

is 24 and now let's divide both sides by

8 so 8x divided by 8 is simply X 24

divided by 8 is 3 and therefore this is

the answer X is equal to 3

and now it's your turn go ahead and work

on this problem let's say about 7x plus

2 minus 3x is equal to 26 so pause the

video and try that so let's begin by

combining like terms let's combine 7x

and negative 3x 7 minus 3 is positive 4

so we have positive 4x plus 2 which is

equal to 26 next let's subtract both

sides by 2 26 minus 2 that's going to be

24 and let's bring down the 4 X now we

need to separate 4 from X so let's

divide both sides by 4 4x divided by 4

is X and 24 divided by 4 is 6 and so

that's the answer X is equal to 6 now

sometimes you may have a variable found

on both sides of the equation so what do

you think we need to do it in is from in

this problem what you want to do is you

remove all the extra variables to one

side of the equation and all the numbers

to the other side so let's move the 5x

from the left to the right side so we

can accomplish that by subtracting both

sides by 5 now simultaneously we can add

8 to both sides we can move all the

constant terms all the numbers to file X

to the left side the 8 will cancel and

the 5x will cancel so on the Left we

just have a number 4 plus 8 is 12 on the

right side we have a variable 8x minus 5

x is 3 X so now all we need to do is

divide by 3 at this point and so X is

equal to 4 and that's it so that's how

you can solve that equation here's a

similar example 13 minus 2x is equal to

4x so now's your turn go ahead and work

on that problem so now what I'm going to

do is add 2x

both sides and at the same time I'm

going to add five to both sides doing it

this way will allow me to get rid of all

the extra variables on the left and all

the constant terms on the right so on

the left I just have 13 plus 5 which is

18 and on a right 4x plus 2x which is 6x

so now all I need to do is divide both

sides by 6 and so X is going to be 18

divided by 6 which is stream sometimes

you may have parentheses in the equation

like this one let's say that 3 times 2x

minus 4 plus 1 is equal to 7 so what do

you think needs to do in order to solve

this particular equation what ideas do

you have well we need to use the

distributive property we move to

distribute 3 to 2x minus 4 so first

let's multiply 3 times 2x that's going

to be 6x next let's multiply 3 by

negative 4 and that's negative 12 let's

rewrite everything else now our next

step is to combine like terms negative

12 and 1 are like terms on the same side

so negative 12 plus 1 we can exchange

that for negative 11 since they're

equivalent to each other now all we need

to do at this point is first add 11 to

both sides

and then we could divide right now we

have 6x which is equal to 7 plus 11

which is 18 and then we could divide by

6 18 divided by 6 is 3 and so that's our

answer X is equal to 3

is another problem that's let's try a

similar problem first

go ahead and try this one so let's begin

with the distributive property let's

multiply five by three X just like we

did before and so that's going to be 15

X next we have five times four which we

know to be 20 and then let's rewrite the

other stuff now let's combine like terms

we can add 20 + 2 which we know it's

going to be 22 so we have 15 X + 22

which is equal to 37 now our next step

is to subtract both sides by 22 37 - 22

that's 15 so now all we need to do is

divide both sides by 15 15 divided by 15

is 1 and so that's going to be our

answer

X is equal to 1 now sometimes you may

have parentheses on both sides of the

equation so let's try an example that

illustrates that so let's say we have 2

plus 4 times 3x plus 2 and let's say

that's equal to 3 minus 4 times well

let's change that let's say it's 2 times

5x plus 1 plus 14 so using what you know

go ahead and solve this particular

equation find a value of x so we need to

use distributive property first let's

multiply 4 by 3x and then by 2 4 times

3x is 12x and 4 times positive 2 that's

equal to 8 on the right side we need to

multiply 2 by 5x which is 10 X and then

2 times 1 which is 2 now let's combine

like terms on the left side we can

combine 2 & 8

on the right side 2 and 14 now 2 plus 8

adds up to 10 and 2 plus 14 is 16 so now

let's isolate the X variables let

subtract both sides by 10 X so doing

this will give us an X variable only on

the left side and let's subtract both

sides by 10 so that we're going to have

a constant number on the right side 12

12 X minus 10 X is 2 X 16 minus 10 is 6

now all I need to do is divide by 2 6

divided by 2 is 3 and so X is equal to 3

and that's the answer is this a new

problem that you could try let's say

that 5 minus 2 times 3 X plus 4 is equal

to 3 minus 4 times 2 X plus 1 so based

on the last problem go ahead and try

these forms so let's start with the

distributive property let's multiply

negative 2 by 3 X that's going to be

negative 6 X and then let's multiply

negative 2 by 4 which is negative 8 on

the right side let's start with negative

4 times 2 X that's negative 8x and then

negative 4 times 1 which is negative 4

now let's combine like terms so we can

combine those two on the left and 3 and

negative 4 on the right 5 minus 8 that's

going to be negative 3 and 3 minus 4 is

negative 1 now let's add 8x to both

sides this will give us an X variable on

the left side and it's going to be

positive and let's add 3 to both sides

so we can get a constant term on the

right side

negative six x plus eight X is 2x

negative one plus three is two so if we

divide both sides by two we can see that

X is going to be one and that's it for

that problem now sometimes you may have

multiple fractions we covered an example

where we only had one fraction in the

equation but here's an example with two

fractions now there's different ways in

which you could solve it you could try

to add one to both sides first to

simplify the equation or you could just

from the beginning eliminate all

fractions which sometimes I'd like to do

notice that we have a denominator of 3

and 2 what is the common multiple of 2 &

3 a common multiple of 2 & 3 is 6

so if we multiply everything by 6 we can

get rid of all fractions so let's

multiply 6 by 2 X divided by 3 6 times

2x is 12 X 12 X divided by 3 well that's

going to be 4x so 2/3 of 6 is 4 next

let's multiply 6 by 4 6 times 4 is 24

and then let's multiply 6 by 3 X divided

by 2 6 times 3x is 18 X divided by 2

that's a 9 X and then 6 times negative 1

is negative 6 and now this looks similar

to equations that we've solved before so

let's begin by subtracting both sides by

4 X and simultaneously let's add 6 to

both sides

twenty four plus six is thirty nine X

minus four X is 5x now all we need to do

is divide both sides by five thirty

divided by 5 is 6 so X is equal to 6 and

that's the answer for that problem here

is a similar problem but this time it's

going to be three fractions in the

equation so go ahead and try this from

now let's find the least common multiple

of 4 & 5 a quick way to find just a

common multiple 4 & 5 is to multiply the

two numbers 4 times 5 is 20 so if we

multiply every fraction by 20 we can

completely eliminate the entire equation

of all fractions so let's do that let's

multiply 20 by 3 X divided by 4 now 20

times 3x is 60 X 60 X divided by 4 is 15

X instead of multiplying first and then

dividing you could divide first again

multiply if we did 20 divided by 4

that's 5 5 times 3x will give us the

same answer of 15 X now what about

two-fifths of 20 well if we take 20

divided by 5 that will give us 4 and

then 4 times 2 is 8 so two-fifths of 20

is 8 now if we multiply 83 over 20 by 20

the tourneys will cancel giving us 83

and now let's solve let's subtract both

sides by 8 83 minus 8 is 75 and that all

we need to do is divide both sides by 15

75 divided by 15 is 5 and so that's the

answer so sometimes you may have to deal

with decimals let's say we have point

two x plus 0.3 which is equal to one

point five now you can go ahead and

solve it or you can eliminate all

decimals now every decimal in this

equation is rounded to the the tenths

place

so to get rid of all decimals all we

need to do is multiply everything by ten

point two x times ten is 2x point three

times 10 is 3 1 point 5 times 10 is 15

and now we could solve it like any other

problem so let's subtract both sides by

3 15 minus 3 is 12 and then we divide by

2 we can see that X is equal to 6 here's

the last example in this video so as you

can see all the numbers are rounded to

the hundreds place so what do you think

we need to do in this problem let's

multiply everything by a hundred so

basically you just need to move the

decimal 2 units to the right 100 times

point 28 X is 28 X point 14 times 100 is

14 point 13 x times 100 is 13 X and

point 74 times 100 is 74 and now we just

got to solve let's subtract both sides

by 13 X and also by 14 so 28 minus 13 is

1574 minus 14 is 16 60 divided by 15

that's equal to 4 and that's the answer

for this problem now I want to show you

one of my algebra courses that might be

useful to you if you ever need it so go

to udemy.com

now in a search box just type in algebra

and it should come up so it's the one of

the image with the black background so

if you select that option and if you

decide to go to course content you can

see what's in this particular course so

the first section basic arithmetic for

those of you who want to focus on

addition subtraction multiplication and

division and it has a video quiz at the

end it's a multiple choice video quiz

you can pause it work on the problems

and see the solutions they covers long

division multiplying two large numbers

and things like that the next tutorials

on fractions add in subtracting

fractions multiplying dividing fractions

converting fractions into decimals and

so forth so you can also take a look at

that next solve the linear equations

which we covered and just more examples

if you need more help with that

the next topic order of operations which

is also useful graphing linear equations

you need to know how to calculate the

slope needs to be familiar with the

slope-intercept form standard form and

just how to tell if lines are parallel

perpendicular and so forth and as a quiz

that goes with that as well the next

topic is on inequalities and absolute

value expressions which are also seen a

typical algebra course and then we have

polynomials and that's a long section

and then factoring you just that's not

the topic you need to master and then

system of equations you can solve it by

elimination substitution there's also

word problems as well sometimes you got

to solve equations with three variables

XY and z so that can be helpful

next quadratic equations how to use the

quadratic formula how to graph them how

to convert between standard and vertex

form and then you have rational

expressions

and radical expressions solving radical

equations simplifying it things like

that and every section has a quiz so you

can always review what you've learned if

you have a test of X day so here we have

complex and magic numbers you need to

know how to simplify those exponential

functions logs I have a lot of videos on

logs and then just this is just

functions in general the vertical lines

has horizontal lines has had itself or

functions even or odd as an conic

sections graph in circles hyperbolas

ellipses parabolas and things like that

there's two video quizzes because it's

actually a long section and finally a

rhythm taking geometric sequences and

series so that's my algebra course if

you want to take a look at it and let me

know what you think