# Solving Inequality Word Questions

*(You might like to read Introduction to Inequalities and Solving Inequalities first.)*

In Algebra we have "inequality" questions like:

### Sam and Alex play in the same soccer team.

Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals.

What are the possible number of goals Alex scored?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

## Turning English into Algebra

To turn the English into Algebra it helps to:

- Read the whole thing first
- Do a sketch if needed
- Assign
**letters**for the values - Find or work out
**formulas**

We should also write down **what is actually being asked for**, so we know where we are going and when we have arrived!

The best way to learn this is by example, so let's try our first example:

### Sam and Alex play in the same soccer team.

Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals.

What are the possible number of goals Alex scored?

Assign Letters:

- the number of goals Alex scored:
**A** - the number of goals Sam scored:
**S**

We know that Alex scored 3 more goals than Sam did, so: **A = S + 3**

And we know that together they scored less than 9 goals: **S + A < 9**

We are being asked for how many goals Alex might have scored: **A**

#### Solve:

**S + A < 9**

**S + (S + 3) < 9**

**2S + 3 < 9**

**2S < 9 − 3**

**2S < 6**

**S < 3**

Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals.

Alex scored 3 more goals than Sam did, so **Alex could have scored 3, 4, or 5 goals**.

Check:

- When S = 0, then
**A = 3**and S + A = 3, and 3 < 9 is correct - When S = 1, then
**A = 4**and S + A = 5, and 5 < 9 is correct - When S = 2, then
**A = 5**and S + A = 7, and 7 < 9 is correct - (But when S = 3, then A = 6 and S + A = 9, and 9 < 9 is incorrect)

## Lots More Examples!

### Example: Of 8 pups, there are more girls than boys.

How many girl pups could there be?

Assign Letters:

- the number of girls:
**g** - the number of boys:
**b**

We know that there are 8 pups, so: g + b = 8, which can be rearranged to

We also know there are more girls than boys, so:

We are being asked for the number of girl pups: **g**

Solve:

**g > b**

**b = 8 − g**, so:

**g > 8 − g**

**g + g > 8**

**2g > 8**

**g > 4**

So there could be 5, 6, 7 or 8 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, and the question isn't clear on that point (sometimes questions are like that).

Check

- When g = 8, then b = 0 and g > b is correct (but is b = 0 allowed?)
- When g = 7, then b = 1 and g > b is correct
- When g = 6, then b = 2 and g > b is correct
- When g = 5, then b = 3 and g > b is correct
- (But if g = 4, then b = 4 and g > b is incorrect)

A speedy example:

### Example: Joe enters a race where he has to cycle and run.

He cycles a distance of 25 km, and then runs for 20 km. His average running speed is half of his average cycling speed.

Joe completes the race in less than 2½ hours, what can we say about his average speeds?

Assign Letters:

- Average running speed:
**s** - So average cycling speed:
**2s**

Formulas:

- Speed = \frac{Distance}{Time}
- Which can be rearranged to: Time = \frac{Distance}{Speed}

We are being asked for his average speeds: **s** and **2s**

The race is divided into two parts:

#### 1. Cycling

- Distance = 25 km
- Average speed = 2s km/h
- So Time = \frac{Distance}{Average Speed} = \frac{25}{2s} hours

#### 2. Running

- Distance = 20 km
- Average speed = s km/h
- So Time = \frac{Distance}{Average Speed} = \frac{20}{s} hours

Joe completes the race in less than 2½ hours

- The total time < 2½
- \frac{25}{2s} + \frac{20}{s} < 2½

Solve:

**\frac{25}{2s} + \frac{20}{s} < 2½**

**25 + 40 < 5s**

**65 < 5s**

**13 < s**

**s > 13**

So his average speed running is greater than 13 km/h and his average speed cycling is greater than 26 km/h

In this example we get to use two inequalities at once:

### Example: The velocity **v** m/s of a ball thrown directly up in the air is given by **v = 20 − 10t**, where **t** is the time in seconds.

At what times will the velocity be between 10 m/s and 15 m/s?

Letters:

- velocity in m/s:
**v** - the time in seconds:
**t**

Formula:

**v = 20 − 10t**

We are being asked for the time **t** when **v** is between 5 and 15 m/s:

Solve:

**10 < 20 − 10t < 15**

**10 − 20 < 20 − 10t − 20 < 15 − 20**

**−10 < −10t < −5**

**−1 < −t < −0.5**

**1 > t > 0.5**

number first, so swap over:

**0.5 < t < 1**

So the velocity is between 10 m/s and 15 m/s between 0.5 and 1 second after.

And a reasonably **hard** example to finish with:

### Example: A rectangular room fits at least 7 tables that each have 1 square meter of surface area.
The perimeter of the room is 16 m.

What could the width and length of the room be?

Make a sketch: we don't know the size of the tables, only their area, they may fit perfectly or not!

Assign Letters:

- the length of the room:
**L** - the width of the room:
**W**

The formula for the perimeter is **2(W + L)**, and we know it is 16 m

- 2(W + L) = 16
- W + L = 8
- L = 8 − W

We also know the area of a rectangle is the width times the length: **Area = W × L **

And the area must be greater than or equal to 7:

- W × L ≥ 7

We are being asked for the possible values of **W** and **L**

Let's solve:

**W × L ≥ 7**

**W × (8 − W) ≥ 7**

**8W − W**

^{2}≥ 7**W**

^{2}− 8W + 7 ≤ 0This is a quadratic inequality. It can be solved many way, here we will solve it by completing the square:

**−**7 to the right side of the inequality:

**W**

^{2}− 8W ≤ −7**W**

^{2}− 8W + 16 ≤ −7 + 16**(W − 4)**

^{2}≤ 9**−3 ≤ W − 4 ≤ 3**

Yes we have two inequalities, because **3 ^{2} = 9** AND

**(−3)**

^{2}= 9**1 ≤ W ≤ 7**

So the width must be **between 1 m and 7 m** (inclusive) and the length is **8−width**.

Check:

- Say W = 1, then L = 8−1 = 7, and A = 1 x 7 = 7 m
^{2}(fits exactly 7 tables) - Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m
^{2}(7 won't fit) - Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m
^{2}(7 fit easily) - Likewise for W around 7 m