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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
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