# 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 ?

Reading: Solving Inequality Word Questions

The flim-flam is to break the solution into two parts :

Turn the English into Algebra .

then practice 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 besides write down **what is actually being asked for**, so we know where we are going and when we have arrived !

The best manner to learn this is by example, so let ‘s try our first case :

### 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?

arrogate 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 in concert they scored less than 9 goals : **S + A < 9**

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

#### Solve:

Start with:

**S + A < 9**

A = S + 3, so:

**S + **

(S + 3)

** < 9**

Simplify:

**2S + 3 < 9**

Subtract 3 from both sides:

**2S < 9 − 3**

Simplify:

**2S < 6**

Divide both sides by 2:

**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** .

arrest :

- 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

b = 8 − g

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

g > barn

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

clear :

Start with:

**g > b**

**b = 8 − g**, so:

**g > **

8 − g

Add g to both sides:

**g + g > 8**

Simplify:

**2g > 8**

Divide both sides by 2:

**g > 4**

so there could be 5, 6, 7 or 8 daughter pups .

Could there be 8 daughter pups ? then there would be no boys at all, and the interview is n’t authorize 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 rapid 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 = distance
**Time** - Which can be rearranged to: Time = 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 = distance
**Average Speed**= 25**2s**hours

#### 2. Running

- Distance = 20 km
- Average speed = s km/h
- So Time = outdistance
**Average Speed**= 20**s**hours

Joe completes the slipstream in less than 2½ hours

- The total time < 2½
- 25
**2s**+ 20**s**< 2½

resolve :

Start with:

25**2s**

+

20**s**

** < 2½**

Multiply all terms by 2s:

**25 + 40 < 5s **

Simplify:

**65 < 5s**

Divide both sides by 5:

**13 < s**

Swap sides:

**s > 13**

So his median rush run is greater than 13 kilometers per hour and his average focal ratio 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 prison term **t** when **v** is between 5 and 15 m/s :

10 < five < 15
10 < 20 − 10t < 15
clear :

Start with:

**10 < 20 − 10t < 15**

Subtract 20 from each:

**10 **

− 20

< 20 − 10t

− 20

< 15

− 20

Simplify:

**−10 < −10t < −5**

Divide each by 10:

**−1 < −t < −0.5**

Change signs and reverse inequalities:

**1 **

>

t

>

** 0.5**

It is neater to show the smaller

number first,

so

swap over:

**0.5 < t < 1**

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

And a sanely **hard** exercise 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 do n’t know the size of the tables, lone their area, they may fit absolutely 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 thousand

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

We besides 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 potential values of **W** and **L**

Let ‘s clear :

Start with:

**W × L ≥ 7**

Substitute L = 8 − W:

**W × **

(8 − W)

** ≥ 7**

Expand:

**8W − W2 ≥ 7**

Bring all terms to left hand side:

**W2 − 8W + 7 ≤ 0**

This is a quadratic inequality. It can be solved many way, here we will solve it by completing the squarely :

Move the number term **−**7 to the right side of the inequality:

**W2 − 8W ≤ −7**

Complete the square on the left side of the inequality and balance this by adding the same value to the

right side of the inequality:

**W2 − 8W + 16 ≤ −7 + 16**

Simplify:

**(W − 4)2 ≤ 9**

Take the square root on both sides of the inequality:

**−3 ≤ W − 4 ≤ 3**

Yes we have two inequalities, because **32 = 9** AND **(−3)2 = 9**

Add 4 to both sides of each inequality:

**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 m2 (fits exactly 7 tables)
- Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m2 (7 won’t fit)
- Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m2 (7 fit easily)
- Likewise for W around 7 m