Question

Given that `S+12>=20color(white)("d")` determine `color(white)("d")"a) "S;color(white)("d")" b) "S+22` and `"c) "S-8color(white)("d")`?

Answer 1

a) `" "s >=8`

b) `s+22 >=30`

c) `s-8>=0`

Explanation:

Treat inequalities in the same way as equations unless you multiply or divide by a negative number, in which case the values stay where they are and the inequality sign changes around.

In a) you are asked to solve for `s`

`s+12-12 >=20-12" "larr` subtract 12 from both sides

`" "s >=8`

In b), the right hand side has to become `30`.

You can achieve this by adding `10` to both sides.

`" "s +12 >= 20`

`s +12 color(blue)(+10) >= 20color(blue)(+10)" "larr` simplify both sides

`s+22 >=30" "larr` here is the `30` you needed.
`" "uarr`

In c), you need to have `-8` on the left side.
Start with the simplified inequality from a)

`" "s >= 8" "larr` subtract `8` on both sides

`s color(red)(-8) >= 8color(red)(-8)" "larr` simplify

`s color(red)(-8) >= 0" "larr` here is the `8` you want
`color(white)(.x)uarr`

Answer 2

A) S `>=` 8

B) S + 22 `>=` 30

C) S - 8 `>=` 0

Explanation:

A) We know by subtracting 12 from both sides:
s + 12 `color(red)(-12)` `>=` 20 `color(red)(-12)`

We get S `>=` 8

B) Here, S + 12 `>=` `color(red)(30)`
Since we already know the value must be `>` 20 to give us our 30 on the right, we can add 10 to both sides, giving us:

S + 22 `>=` 30

C) Finally, using the logic from how we did A, by default for either solution in A or B to work, S - 8 would have to be initially greater than (or equal) to zero. Therefore:

S - 8 `>=` 0

Answer 3

Once you get used to this type of question structure the solution derivation is very quick. A lot of explanation given.

Explanation:

`color(blue)("The teaching bit")`

Before we start you must understand that there is big difference between 'bigger' and greater also smaller and less.

Bigger and smaller is looking at something that has a special name. Once said you can forget it as at your level of maths the name will not crop up. The name is 'Cardinality'.

Example: -36 is just as 'big' as +36. It is the count from zero that is the measure. The fact that one is negative and the other is positive does not come into it for Cardinality (size).

Given that right of 0 on the number line is positive:

'Less than' is to the left of what ever value you are looking at.
'More than' is to the right of whatever number you are looking at.

Example: `-1/4` is to the right of `-1/2` so; `-1/4` is greater than `-1/2`

Had you twigged that the number line is the same sort of thing as the x-axis in a graph.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering your question b")`

On the whole you may treat inequalities like equations. However, you must keep in mind what it is actually saying.

`color(brown)("Given that "s+12>=20`

Comparing the given to the objective the s is the same. However, the right hand sides are not the same. So we need to 'force' the right hand side to change from 20 to 30

Add `color(red)(10)` to both sides

`color(green)(s+12>=20 color(white)("dd")->color(white)("dd")s+12color(red)(+10)>=20color(red)(+10))`

`color(white)("ddddddddddd")color(green)(->color(white)("d")color(white)("dd")s+22color(white)("d.d")>=30`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering your question c")`

`color(brown)("Given that "s+12>=20`

We to 'force' the 12 to change to 8

`12-4=8`

Subtract `color(red)(4)` from both sides

`color(green)(s+12>=20 color(white)("dd")->color(white)("dd")s+12color(red)(-4)>=20color(red)(-4))`

`color(green)(color(white)("dddddd.ddddd")->color(white)("dddd")s+8color(white)("d")>=16)`