Hey guys,
welcome to Algebra I.
Today's lesson focuses
on solving multi-step linear
inequalities algebraically.
Your knowledge
of solving equations
and properties of inequality
helps you move through
this lesson easily.
You ready to get started?
Let's go.
Okay, so, before we jump
into inequalities,
I want to throw back
to equations.
They're not--
There obviously are great
differences between them,
but steps you take
to solve an equation
and to solve an inequality
algebraically are very similar.
You can do one,
most likely,
you can do the other.
Let's look here:
6x minus 3 equals 9.
If I solve this
to get the solution,
like they're
asking me to,
I'd start by adding 3
to both sides, right?
Cancel out right there,
so 6x equals 12.
Divide both sides by 6.
x is 2.
What that means is
the solution to this equation,
the value when
I substitute it,
giving the equation
a value of 9, is 2.
Let me show you
what I mean, I'll verify.
My equation
is 6x minus 3 equals 9.
We're saying 2 is the only
solution to this equation.
Let's substitute
in there and check--
6 times 2 minus 3.
6 times 2, that's 12.
So, 12 minus 3 is 9,
so 2 is the only solution
to this equation.
Let's move on a little.
We'll change that equation
to an inequality.
I want to show the difference
between an equation
and an inequality.
As far as solving this,
it's essentially
the same thing.
Add 3 to both sides, okay.
That's going to cancel out.
So 6x is less than 12.
Divide both sides by 6.
x is less than 2.
This time, we were asked
for the solution set
because in an inequality,
you always have more
than one solution.
There are infinitely
many solutions
that will prove
your inequality is true.
What this is saying is,
any x value that's less than 2
will make
this inequality true.
Let me show you.
Let's pick numbers
less than 2.
Let's pick 1--
1 is less than 2.
Split this up.
We have 6x minus 3
is less than 9,
so if we let x be 1:
6 times 1 minus 3...
6 times 1 is 6,
and 6 minus 3 is 3.
So 3 is less than 9,
so 1 works.
What if x were 0?
Let's try that:
6 times 0 minus 3--
6 times 0 is 0.
0 minus 3 is negative 3,
which is less than 9,
so that's true.
We could go on for days.
Any x value
that's less than 2
will make
this inequality true.
That's the difference between
the solution to an equation
and a solution set
to an inequality
is there is more
than one value
that will make
that inequality true.
Let's practice solving
inequalities for a bit.
We have 10x minus 8
is greater than 3x plus 20.
We'll get the solution set
and write it
in set-builder notation,
just so you keep
practicing that skill.
Think of your inequality sign
as dividing
that inequality into 2 sides.
We want our "x"s on one side,
numbers on the other,
just like when we
were solving equations.
I subtract 3x from both sides,
my first step.
You might have taken another.
That's perfectly fine.
I have 10x minus 3x,
so that's 7x minus 8
greater than--
3x minus 3x,
that wipes out.
I'm bringing down the 20.
So, now I'm going
to add 8 to both sides
because I'm trying
to isolate that 7x.
So, plus 8, plus 8.
That's wiped out.
So, 7x is greater than 28.
Scroll down a little bit.
You're at your last step.
What do you need to do
to get that x by itself?
Let's divide both sides by 7,
so x is greater than,
and 28 divided by 7 is 4.
If we write this
in set-builder notation,
it looks different
for inequalities.
Looks like this.
We have our brace.
You have an x,
you have a colon,
then you have your x
is greater than 4,
then you close your brace.
So with how
you read this,
you read it like this:
"x such that
x is greater than 4."
That colon you read
as "such that."
I know it sounds very mathy,
for lack of a better term,
because we don't typically
speak that way,
but that
is set-builder notation.
When you see the colon,
you read that as "such that."
This is, "x such that x
is greater than 4."
And that's the solution
to this inequality.
Let's try another one.
We have 5 halves x plus 30
is greater than
or equal to 10.
They're asking us to use
set-builder notation.
I'm moving this problem
to the middle here,
just so it's easier
for me to work out.
You saw that fraction.
Fractions typically alarm us,
but we're good.
We know how to handle it.
We'll clean that up.
Now let's get
that x term by itself.
Let's start by subtracting
the 30 from both sides.
All right.
That wipes out right there.
I'm going to bring down
that 5 halves x
greater than
or equal to 10 minus 30.
That's negative 20.
Now treat that fraction as if
this is a division problem:
5 divided by 2.
So, the opposite
of dividing by 2,
the inverse operation,
is to multiply by 2.
I multiply the left side by 2
and the right side by 2.
On the left side,
those 2s are going to cancel.
I'm left with 5x
greater than or equal to
negative 20 times 2.
That's negative 40.
We're at
the very last step.
What do we need to do here
to get x by itself?
Divide both sides by 5.
So, x is greater than
or equal to negative 8.
If I wanted to represent
that answer
using set-builder notation--
get a little more space here.
We write that as--
put our brace--
x--the colon
for the "such that"--
x is greater than
or equal to negative 8.
That's how we represent
that solution
using set-builder notation.
Any value of x that is greater
than or equal to negative 8
proves this
inequality is true,
the one we started
with up there.
Let's keep moving.
Let's do another one together.
We represent the solution
using set-builder notation.
I'll move that to the center.

(female describer)
4 times x plus 1
minus 9x is less than 54.

(teacher)
Okay, let's first use
the distributive property
so we can get this
cleaned up a bit.
So, 4 times x--4x--
4 times 1--plus 4--
minus 9x--
going to bring everything
else down--less than 54.
We have like terms
on the left side we combine:
the 4x
and the negative 9x.
You combine 4x
and negative 9x.
That's negative 5x
plus 4, less than 54.
Okay.
Going to keep moving along.
Scroll down a little bit.
I'm trying to isolate
that negative 5x,
so let's subtract 4
from both sides.
Bring down our negative 5x,
is less than 54 minus 4.
That's 50.
Give me
a little more room here.
Okay, we're at the last step
to get x by itself.
We need to divide
both sides by negative 5.
This is going to cancel,
and we have x.
Remember that property
of inequality.
If we divide both sides
by a negative number,
we flip that
inequality sign--
changes the relationship.
50 divided by 5,
that's negative 10.
My solution set--
have my brace--
x colon x is greater
than negative 10.
This is, "x such that x
is greater than negative 10."
We got through that one.
And I do believe
it's your turn to try.
Press pause.
Take your time
and work through these.
When you're ready to compare
your answers, press play.

(female describer)
Represent each solution set
using set-builder notation.
1. 2/3x minus 1
is less than or equal to 5.
2. 5x plus 4
is greater than 7x plus 10.

(teacher)
Okay, let's see how you did.
Let's get these little
answer hiders out of the way.
For number one,
you should have got
x such that x is less
than or equal to 9.
For number 2,
x such that x is less than 3.
Let me show you
how I got those,
in case you need to see.
We're going to start out--
make sure I've got the pen.
We'll isolate that 2/3,
so I start
by adding 1 to both sides.
Plus 1, plus 1,
so cancel that right there.
Bring down my 2/3x,
less than or equal to--
5 plus 1 is 6.
I treat this fraction
like it's a division problem.
So 2 divided by 3--
the opposite of dividing by 3
is to multiply by 3.
I'm going to multiply
both sides by 3.
On the left side,
cancel those 3s out
and I'm just left with 2x,
less than or equal
to 6 times 3.
That's 18.
Little more workspace.
Then for our last step,
divide both sides by 2,
so x is
less than or equal to--
18 divided by 2 is 9.
If we represented that
in set-builder notation--
got our brace--
x such that
x is less than
or equal to 9, okay?
Let's try the other one.
So, want to isolate x.
I'm going to start
by subtracting
7x from both sides.
That cancels
on the right.
5x minus 7x,
that's negative 2x
plus 4 greater than 10.
Next step's going to be
to subtract 4 from both sides.
So, minus 4, minus 4.
All right.
So, negative 2x,
greater than--
10 minus 4 is 6.
Final step--
divide by negative 2.
Because we're dividing
both sides by a negative,
we've got to flip
that inequality sign.
6 divided by 2
is negative 3.
If we wrote that using
set-builder notation--
let's get a little
more space underneath--
brace, x such that x
is less than negative 3.
Now just before
I let you go,
another way you see
set-builder notation--
instead of a colon,
you may see just a bar,
like, "x such that x
is less than negative 3."
That's another way
you may see the set-builder
notation, okay?
Great job, guys.
Now you're familiar
with how to solve
multi-step linear
inequalities algebraically.
Hope to see you
for more Algebra I. Bye.
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