welcome to the Cypress College math review on linear inequalities
in this video you are going to learn to solve linear inequalities and express
the solution in various forms in objective one you will learn to write
interval and set builder notation
given the inequality X is greater than two there are different ways to writing
a solution to this inequality first note that X is greater than two means that X
can be three four five six ten a thousand a million a zillion
furthermore X can also be two point one because two point one is greater than
two 2 point 2, 2 point 3, 2 point three three, 2 point three, 2 four so
basically there are many numbers that are greater than two so we can represent
the solutions of this inequality by graphing it on a real number line
so here is a real number line where we mark the negative values to the left of
zero and the positive values to the right of zero since X is greater than 2
then X can be any numbers to the right of 2 and we graph it by shading the
region to the right of 2 another representation of this
inequality is interval notation note that on the real number line the numbers
go infinitely to the right where we represented by positive infinity and on
the left the numbers go infinitely to the left and that is represented by
negative infinity we look at the shaded region on the
graph and see that it starts from 2 and it goes infinitely to the right to
positive infinity therefore we write 2 comma infinity the number on the left is
where the shaded region on the graph starts from and the number on the right
is where it ends the third representation of this
inequality is the set builder notation this is read
the set of all X such that X is greater than 2 this bar here means such that
note that what goes after the bar is exactly the given inequality
here is the quick chart to help you distinguish between the three types of
writing the solutions to the inequality
suppose X is the variable and a is some arbitrary number if we have X is greater
than a then we graph by first marking a on the number line and shade to the
right of a because greater implies shading to the right so the graph goes
infinitely to the right or positive infinity then based on this graph we get
the interval notation as I mentioned earlier we look at the graph from left
to right it starts from a and it goes to infinity therefore in the interval
notation the left endpoint is a and the right endpoint is infinity and for the
set builder notation we have the set of all X such that X is greater than a
recall that this bar means such that and what goes after the bar is exactly the
given inequality here we have X is greater than or equal to a so we still
mark a on the line and shade to the right of a because X is greater than so
the graph goes infinitely to the right or to positive infinity to indicate that
X can equal to a we put a bracket at a so the bracket means the endpoint is
included note in the previous one the inequality does not include equal two
which means X cannot be a so we use a parenthesis around a to indicate that a
is not included now we get the interval notation of this
by looking at the graph the shaded region starts from a on the left and
ends at infinity on the right so we have a comma infinity note that we use a
bracket around a to indicate that a is included we always use a parenthesis
around infinity or negative infinity because infinity is not a number but
rather a notation that indicates a large quantity
for set-builder notation once again note that what goes after the bar is exactly
the given inequality
here we have X is less than a less than means shading to the left so we mark a
on the number line and shade to the left of a so the graph goes infinitely to the
left or to negative infinity since X is strictly less than a we use a
parenthesis on a then based on this graph we get the interval notation on
the left side of the shaded region we have negative infinity and the shaded
region ends up a and of course we use parentheses around a because X does not
equal to a
for the set-builder notation once again what goes after the bar is exactly the
given inequality
here we have X is less than or equal to a so this means a is included since X is
less than a then we shade to the left of a and it goes infinitely to negative
infinity also since X is equal to a we use a bracket on a then for the interval
notation we once again based on the graph on the Left we have negative
infinity and it ends at a for the set builder notation once again what goes
after the bar is exactly the given inequality
so there are four things you need to pay attention to if the variable is greater
than a number then we shade to the right of that number if the variable is less
than or less than or equal to a number then we shade the left side of the
number if the inequality does not include equal to then we use a
parenthesis around the number but if the inequality sign includes the equal to
then we use bracket and always remember to use parentheses around infinity and
negative infinity
let's look at this example we want to graph the inequality and then write the
solutions in interval notation and in set-builder notation for x is less than or
equal to negative 4 we mark negative 4 on the number line since x is less than
negative 4 that means X is negative 5 negative 6 negative 7 so on and so forth
which is the region left of negative 4
since the inequality sign has equal 2 then we use a bracket at negative 4
then we look at the graph to get the interval notation the graph goes
infinitely to negative infinity so on the left of the shaded region we have
negative infinity and on the right side we have negative for around negative 4
we use a bracket around negative infinity we always use a parenthesis now
for set builder notation we have the set of all X such that that's the bar X less
than or equal to negative 4
in this example we have X is greater than five we want to graph the
inequality then write the solutions in interval notation and in set-builder
notation we begin by marking five on the number line
since x is greater than five means x is six seven eight nine ten and even the
decimal values between those points that means we shade to the right of five and
since the inequality symbol does not have the equal sign then we use a
parenthesis around five so this graph goes infinitely to the right or positive
infinity you don't need to put infinity here like I did but I just want to write
it there to help you see when we write the interval notation so based on the
graph we have the left value is five and the right is infinity since we use
parenthesis around five in the graph we also use parenthesis in the interval
notation and as always we use parentheses around infinity so remember
that when you write the interval notation that does not include the
endpoint use parentheses for set-builder notation we have the set of all X such
that X is greater than five
pause the video and try these problems
objective two solving linear inequalities with variables on one side when finding
the solutions to an inequality solve for the variable like you would as if it
were an equation just remember when you divide or multiply by a negative number
you need to flip the inequality let's look at a few examples for
instance if you have negative 5 X less than 15 then to solve for X we would
divide by negative 5
negative 5 divided by negative 5 is 1 so we have X 15 divided by negative 5 is
negative 3 and since we divided by a negative number then this inequality
sign is flipped to greater than
another example negative 2 X is greater than or equal to negative 10 so we would
divide by negative 2 to get X by itself we get X on the Left negative 10 divided
by negative 2 is positive 5 and since we're dividing by a negative number then
we switch the inequality sign and this becomes less than or equal to
another example 4x is less than negative 12 so we divide by 4 we get X on the
left negative 12 divided by 4 is negative 3 however we are not dividing
by a negative number so the inequality remains as is so you
see that it does not matter what the end value is it could be positive or
negative the only time we flip the inequality is
when we divide or multiply by a negative number
let's solve for x in this inequality then express the solution in the three
forms that we just learned so we would solve for this as if it were an equation
we want to isolate X so we first need to add 8 to both sides
negative eight plus eight is zero so we are left with 2x negative four plus
eight is four we bring down the inequality now we divide by two
2 divided by 2 is 1 so we have X on the left 4 divided by 2 is positive 2 since
we are not dividing by a negative number so the inequality remains the same x is
greater than 2 to graph this we marked you on the
number line and just for the purpose of convenience we do not need to label more
numbers than two X is greater than 2 so which side do you
think we need to shade
yes the right side
now do we use a bracket or parentheses
parentheses yes that's right because there isn't an equal in the inequality
please try and write the interval notation yourself
it should be two comma infinity
and the set builder notation is the set of all X such that X is greater than 2
once again what goes after the bar is the inequality there
to solve for P in this inequality we first need to combine the like terms
negative 3p minus 4p is negative 7p
then we divide both sides by negative seven
negative seven divided by negative seven is just one so we have P on the left
hand side 28 divided by negative 7 is negative 4
since we're dividing by a negative number then we must flip the inequality
so it becomes less than or equal to now we graph this by marking negative 4 on
the number line and which size should we shade
yes the left side because P is less than less than or less than or equal to means
shading to the left do we use a bracket or parentheses here yes we use a bracket
because there is an equal in the inequality
please pass the video and write in the interval and set builder notation
you should get negative infinity comma negative for bracket at negative 4
because we use a bracket in the graph this also means that negative 4 is
included for the set builder notation we have the set of all P since the variable
used in the problem is P such that we copy this P is less than or equal to
negative 4
to self okay we need to combine 5k with 4k first that gives 9k
now we divide both sides by 9 to get K by itself
negative 18 divided by 9 is negative 2 9 K divided by 9 is K
do we flip the inequality no because we're not dividing by a negative number
now we should always repay first although it is written on the right hand
side this is actually K it's greater than negative 2 or to make it simple we
write K on the left so we have K is greater than negative 2
notice the inequality opens towards K so when you change K to the left the
inequality should still be opening towards K the reason that I recommend
that you write k on the left is because it makes it easy to graph and to write
the set builder notation
we mark negative two on the number line since K is greater than we shade to the
right of negative two since there is an equal in the
inequality then we use parentheses then for the interval notation we have
the shaded region begins from negative two and it goes infinitely to positive
infinity we use parentheses around negative two
and for the set-builder notation we have the set of all K since the variable used
in the problem is K such that K is greater than negative 2
pass the video and try these problems
you
objective three solving linear inequalities with variables on both
sides example soft graph the solution set and
write the solution in interval and set builder notation for this inequality
notice we have the variable written on both sides of the inequality so to solve
for X we first need to gather all the X terms to one side so we're going to
subtract 2x from both sides
negative 5x minus 2x is negative 7x
two X minus two x is 0 so we are left with 1 then we add 3 to both sides
negative 3 plus 3 is 0 1 plus 3 is 4 and now we divide by negative 7 to get X by
itself we have X on the right 4 divided by negative 7 is negative 4 over 7 and
it is okay to get a fraction since we are dividing by a negative number we
need to flip this inequality so it becomes less than we have X is less than
negative 4 over 7 another way we can solve for x in this
inequality is to bring X to the right-hand side by adding 5x to both
sides negative 5x plus 5 X is zero so we have negative 3 is greater than 2x plus
5x is 7x
then subtract one from both sides so that the X term is by itself on one side
negative three minus one is negative four
one minus one is zero and now we divide by seven
negative 4 divided by 7 is negative 4/7 and then we get X on the right hand side
since we are not dividing by a negative number the inequality remains the same
comparing the two inequalities we have less than here and greater than over
here however as I have mentioned in objective 2 we should always read the
variable first so this is read X is less than negative 4 over 7 which is
essentially the same as what we got on the Left we can rewrite this to have X
on the left so it should be X is less than negative 4 over 7 so my
recommendation is to always move X to the left hand side so that it is easy to
graph and to write the set builder notation later
now to graph this we mark negative 4/7 on the number line
which side do we shade yes the left-hand side since X is less
than
and do we use a bracket or parentheses at negative 4/7
yes parentheses because X is strictly less than there isn't equal to in the
inequality please pause the video and write the interval and set builder
notation yourself
the interval notation should be negative infinity comma negative 4/7 once again
we look at the shaded part since the graph goes infinitely to negative
infinity then we have negative infinity on the left and it ends at negative 4
servants around negative infinity we use parentheses and since negative 4/7 is
not included we use parentheses which is also what we used in the graph for the
set builder notation we get the set of all X such that X is less than negative
4 servants
notice we have the variable written on both sides of the inequality so to solve
this we need to gather them to one side as I had mentioned before it is more
convenient to move the variable to the left hand side so we are going to
subtract 8 Y from both sides
why - 8 Y is negative 7 y bring down negative 6 and the inequality 8y minus
8y is 0 so we have 15 on the right please pass the video and try to solve
for y yourself
let's go over this we add six to both sides negative six plus six is zero
so we have negative 7y 15 plus 6 is 21 then we divide by negative seven
negative 7y divided by negative 7 is y 21 divided by negative 7 is negative 3
and since we are dividing by a negative number we flip the inequality so we get
Y is greater than or equal to negative 3 let's graph this we mark negative 3 on
the number line and we shade to the right of negative 3 since Y is greater
than an at negative 3 we use a bracket because Y is also equal to negative 3
based on the graph we get the interval notation to be negative three comma
infinity and we put a bracket around negative three for the set builder
notation we get the set of all Y since the variable used in the problem is Y
such that Y is greater than or equal to negative three
and this completes the problem
pause the video and try these problems
you
objective for solving linear inequalities by using the
distributive property
we would solve for this as if it were an equation so to solve for x in this
inequality we first need to distribute five to remove the parentheses so we
take five times seven which gives 35 five times 2x is 10x
then bring down the terms on the right side now we gather the terms involving X
to the left-hand side so we add X to both sides
ten X plus one X is 11 X
negative x plus x is zero so we get 8 on the right hand side
please pause the video and try to finish the problem yourself restart the video
when you are ready to check your answer let's go over this so we want to
subtract 35 from both sides so that we have the variable on one side
and constant on the other 35 - 35 is zero so we get 11 X on the left 8 minus
35 is negative 27 then we divide both sides by 11 to get X by itself
we get X is greater than or equal to negative 27 over 11
notice that the inequality stays the same because we're not dividing by a
negative number now to graph this we mark negative 27 over 11 on the number
line and it's okay to get a fraction whatever the number we get which is
market on the number line we shade to the right side of negative 27 over 11
because X is greater than we put a bracket at this endpoint since X does
equal to negative 27 over 11 then we look at the graph to get the
interval notation it starts from negative 27 over 11 and it goes
infinitely to positive infinity for the set-builder notation we get the
set of all X such that X is greater than or equal to negative 27
over 11
and this completes the problem
the first thing we need to do in this inequality is to distribute to remove
the parentheses there is an invisible negative one in front of this
parenthesis so we have negative 1 times 4x is negative 4x negative 1 times
negative 2 is plus 2
on the right hand side take negative 5 times both terms inside the parentheses
negative 5 times X is negative 5 X negative 5 times 1 is negative 5 then
bring this negative 5 down what do you think we need to do next
yes we need to combine the like terms 8 plus 2 is 10
over here we combine negative five with negative five since there are both
constant and they give negative ten
now we gather the terms involving X to the left-hand side
so we add 5x to both sides
negative 4x plus 5 X is 1x
negative five x plus five x is zero so we get negative ten on the right side
now we subtract 10 from both sides ten minus ten is zero so we get X is
less than negative ten minus 10 is negative 20
we graph this by marking negative 20 on the number line and we shade to the left
of negative 20 since X is less than and we use a parenthesis at negative 20
since X is strictly less than negative 20 the graph goes infinitely to negative
infinity on the left and it ends at negative 20
and the set builder notation is the set of all X such that X is less than
negative 20
for this last example please pause the video and try the problem yourself
restart the video when you are ready to check your answer
let's go over this so you see here we need to first distribute on both sides
so we take 3 times X to get 3x 3 times negative 5 to get negative 15 2 times 2x
is 4x 2 times negative 1 is negative 2 and then we subtract 4x from both sides
so that now we have the X term on the Left
3x minus 4x is negative X bring down the negative 15 4x minus 4x is 0 so we get
just negative 2 on the right now add 15 to both sides
negative 15 plus 15 is zero so we get negative X on the Left negative 2 plus
15 is 13 there is an invisible negative one in
front of X so we divide by negative one
negative X divided by negative 1 is positive X 13 divided by negative 1 is
negative 13 and since we are dividing by a negative number we flip the inequality
so it becomes X is greater than or equal to negative 13
to graph this we mark negative 13 on the number line and we shade to the right
side because X is greater than and we use a bracket at negative 13
we look at the graph to get the interval notation it starts from negative 13 and
it goes infinitely to positive infinity on the right
and finally the set builder notation is the set of all X such that X is greater
than or equal to negative 13
pause the video and try these problems