Video Transcript
In this video, we will learn how to
solve a linear inequality in two steps. We will begin by recalling some key
notation and recall how to solve one-step linear inequalities. We will then look at questions
involving setting up and solving two-step linear inequalities, including those
involving real-world problems. We will start by looking at how we
can write inequalities and what they mean. If we consider the variable π₯ and
the constant three, there are four different ways we can link these using an
inequality sign.
The first inequality means that π₯
is greater than three. The second inequality means that π₯
is less than three. The third inequality means that π₯
is greater than or equal to three. And finally, the bottom inequality
means that π₯ is less than or equal to three. These inequalities can also be
written as intervals. If π₯ is greater than three, it can
take any value above three up to β. When π₯ is less than three, it can
take any value below three down to negative β. Notice here that we have curly
brackets or parentheses as our value can never reach β or negative β, and itβs also
greater than three and less than three. When π₯ is greater than or equal to
three, we use a square bracket alongside the three. This is also true when π₯ is less
than or equal to three.
β and negative β will always have a
curly bracket. We might also be told that π₯ lies
in the interval between four and eight. There is a square bracket by the
four, which means that π₯ is greater than or equal to four. As there is a curly bracket by the
eight, π₯ must be less than eight. This can be written using
inequality signs. π₯ is greater than or equal to four
and less than eight. We can also display this using set
notation, where π₯ contains all the integer values between four and seven
inclusive. As π₯ is less than eight, this is
not included in our set. We will now clear some space to
show how inequalities can be displayed on a number line.
Letβs consider the two examples π₯
is greater than three, and π₯ is greater than or equal to three. If π₯ is greater than three, we
have an open circle at three on the number line. As π₯ can take any value greater
than this, we draw an arrow to the right. The only change when π₯ is greater
than or equal to three occurs at the number three, where we color in the circle or
dot. This indicates that π₯ can now take
the value three or any number to the right.
We will now have a quick reminder
of how we solve one-step inequalities. We can solve one-step inequalities
in the same way as we solve one-step equations. Letβs consider the four
inequalities shown. Firstly, we have π₯ plus three is
greater than five. In order to solve this, we perform
the inverse or reciprocal operation. The opposite of adding three is
subtracting three. Subtracting three from both sides
of the inequality gives us the answer π₯ is greater than two. The inverse or opposite of
subtracting nine is adding nine.
As four plus nine is equal to 13,
the solution to our second inequality is π₯ is less than or equal to 13. Five π₯ means five multiplied by
π₯. The inverse of multiplying by five
is dividing by five. 35 divided by five is equal to
seven. So, π₯ is greater than or equal to
seven. In our final one-step inequality,
we need to multiply both sides by seven. The left-hand side becomes π₯, and
four multiplied by seven is 28. The answer to this inequality is
therefore π₯ is less than 28. We will now move on to solve some
problems involving two-step inequalities.
Find the solution set of three
π₯ minus seven is less than negative four given that π₯ is a natural number.
Before trying to solve this
inequality, it is worth recalling what a natural number is. The natural numbers are the
nonnegative integers, for example, zero, one, two, three, four, and so on. We will now solve the
inequality given and find which of these numbers satisfy the inequality. Our inequality states that
three π₯ minus seven is less than negative four. We can solve this using inverse
operations. Our first step is to add seven
to both sides of the inequality, as the opposite of subtracting seven is adding
seven. Negative four plus seven is
equal to three, so three π₯ is less than three.
Our second and final step is to
divide both sides of this new inequality by three. Three π₯ divided by three is
equal to π₯, and three divided by three is equal to one. The solution to our inequality
is π₯ is less than one. This answer can be written in
interval notation, where π₯ can take any value less than one down to negative
β. In this question, however,
weβre asked for the solution set. π₯ also needed to be a natural
number. The only natural number that is
less than one is zero. This means that the solution
set of the inequality three π₯ minus seven is less than negative four where π₯
is a natural number is the set containing the number zero.
We will now look at a second
example where we need to write the solution set as an interval.
Find the solution set of the
inequality negative two π₯ plus three is less than or equal to five. Write your answer as an
interval.
The first part of this question
will involve solving the inequality using two steps. We will, then, write this
answer as an interval. The inequality negative two π₯
plus three is less than or equal to five can be solved using inverse
operations. Our first step is to subtract
three from both sides of the inequality. As five minus three is equal to
two, we have negative two π₯ is less than or equal to two. Our second and final step is to
divide both sides of the inequality by negative two. We do need to be careful here
as we recall, if negative π₯ is less than four, then π₯ is greater than negative
four.
When dividing an inequality by
a negative number, the sign also changes. Negative two π₯ divided by
negative two is π₯. Two divided by negative two is
negative one. If negative two π₯ is less than
or equal to two, then π₯ is greater than or equal to negative one. This means that π₯ can take any
value greater than or equal to negative one. We were asked to write this as
an interval, so π₯ is greater than or equal to negative one but less than β. The equal to part of the
inequality means that we have a square bracket next to negative one. β and negative β will always
have a curly bracket or parentheses as we can never reach these values.
Our next question will involve
writing an inequality in a practical setting.
A candy store has a special
offer; if you spend more than 15 dollars, you get a free chocolate drink. Gift boxes are three dollars
each, and chocolates are two dollars per 50 grams. Write an inequality to find π€,
the weight of chocolate you must buy with a gift box, if you want to receive a
free chocolate drink.
We know that the special offer
of a free chocolate drink happens if you spend more than 15 dollars. This means that our expression
must be greater than 15. We know that one gift box costs
three dollars. We also know that chocolates
cost two dollars for 50 grams. Dividing both of these by two
means that we can buy 25 grams of chocolate for one dollar. The weight of chocolate we need
to buy is π€. Therefore, the cost of this
will be π€ divided by 25 as each 25 grams of chocolate costs one dollar. Weβre also buying one gift box
which costs three dollars. This means that our total spend
is π€ over 25 plus three.
To receive the free gift, this
must be greater than 15. The inequality to find π€ is
therefore π€ over 25 plus three is greater than 15. Whilst we donβt need to do so
in this question, we could solve the inequality by firstly subtracting three
from both sides. This would give us π€ over 25
is greater than 12. Our second step would be to
multiply this inequality by 25. The inverse or reciprocal
operation of dividing by 25 is multiplying by 25. 12 multiplied by 25 is equal to
300. This means that you would need
to buy more than 300 grams of chocolate to qualify for the free chocolate
drink.
Our next question will involve
setting up and then solving a two-step inequality.
Matthew needs to buy some
clothes. The storeβs parking lot has the
shown sign outside. Parking: the first hour is free, one
dollar 50 per hour after that. Write an inequality for π‘, the
time in hours, that Matthew can park if he only has eight dollars 25 in
cash. Given that you must pay for
whole hours of parking, use your inequality to find the maximum time that
Matthew can park.
Letβs consider the information
given on the sign. Weβre told that the first hour
of parking is free, and the amount of time parked in hours is π‘. Every hour after that costs one
dollar 50. So, we mightβve seen that we
need to multiply one dollar 50 by π‘. However, as that first hour is
free, we need to multiply one dollar 50 or 1.5 by π‘ minus one. We know that parking for two
hours would only cost one dollar 50. And substituting two into this
expression gives us one dollar 50. Likewise, three hours of
parking would cost three dollars as the first hour is free. Substituting three into the
expression gives us three minus one, which is two, and multiplying this by 1.5
gives us three dollars.
Matthew only has eight dollars
and 25 cents in cash. Therefore, this expression
needs to be less than or equal to 8.25. We could distribute the
parentheses on the left-hand side. However, there is no need at
this time. The inequality in terms of π‘
is 1.5 multiplies by π‘ minus one is less than or equal to 8.25. The second part of this
question asks us to solve the inequality to find the maximum time that Matthew
can park. We can solve the inequality
using inverse operations.
Our first step is to divide
both sides by 1.5. The left-hand side of the
inequality becomes π‘ minus one. 8.25 divided by 1.5 is 5.5. Therefore, π‘ minus one is less
than or equal to 5.5. Our second and final step is to
add one to both sides of the new inequality. This gives us π‘ is less than
or equal to 6.5. Weβre told that you must pay
for whole hours of parking. Therefore, π‘ must be an
integer. As π‘ must be less than or
equal to 6.5, the greatest integer value it can take is six. This means that the maximum
time that Matthew can park for is six hours.
We will now summarize some of the
key points from this video in solving two-step linear inequalities. Two-step linear inequalities are
generally written in the form ππ₯ plus π is greater than π. There are four possible signs,
greater than, less than, greater than or equal to, or less than or equal to. The final solution or answer can be
written as an interval or solution set. It can even be shown on a number
line. If π₯ is greater than four and less
than or equal to seven, this can be written as the interval four to seven with a
curly bracket or parenthesis by the four and a square bracket by the seven.
As an integer solution set, this
can be written as the set of numbers five, six, and seven, as we cannot be equal to
four, but we can be equal to seven. When solving the inequality, we
must perform the inverse or reciprocal operation at each step. Addition and subtraction are
inverse operations as are multiplication and division. Remember, whatever you do to one
side of the inequality sign, you must do to the other. Setting up and solving linear
inequalities can help us solve many real-life mathematics problems.