Types of polygons

Solving Inequality Word Questions

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

In Algebra we can 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 Sam might have scored: S

Solve:

Start with:		S + A < 9
A = S + 3, so we can substitute "S + 3" for A:		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.

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

b = 8 − g

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

g > b

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

Solve:

Start with:		g > b
b = 8 − g, so we can substitute "8 − g" for b:		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 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, 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 = 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 = Distance/Average speed = 20/s hours

Joe completes the race in less than 2½ hours

• The total time < 2½
• 25/2s + 20/s < 2½

Solve:

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 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 – 5t, where t is the time in seconds.

At what times will the velocity be between 5 m/s and 15 m/s?

Letters:

• velocity in m/s: v
• the time in seconds: t

Formula:

• v = 20 – 5t

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

5  <  v  <  15

5  <  20 − 5t  <  15

Start with:		5  <  20 − 5t  <  15
Subtract 20 from each:		5 − 20  <  20 − 5t − 20  <  15 − 20
Simplify:		−15  < −5t  <  −5
Divide each by 5:		−3  < −t  <  −1
Change signs and reverse inequalities:		3  >  t  >  1
It is neater to show the smaller number first, so:
Swap over (for a neater answer):		1  <  t  <  3

So the velocity is between 5 m/s and 15 m/s between 1 second and 3 seconds 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:

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 bycompleting the square:

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 −32 = 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

Solving Inequality Word Questions

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

In Algebra we can 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 Sam might have scored: S

Solve:

Start with:		S + A < 9
A = S + 3, so we can substitute "S + 3" for A:		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.

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

b = 8 − g

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

g > b

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

Solve:

Start with:		g > b
b = 8 − g, so we can substitute "8 − g" for b:		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 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, 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 = 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 = Distance/Average speed = 20/s hours

Joe completes the race in less than 2½ hours

• The total time < 2½
• 25/2s + 20/s < 2½

Solve:

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 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 – 5t, where t is the time in seconds.

At what times will the velocity be between 5 m/s and 15 m/s?

Letters:

• velocity in m/s: v
• the time in seconds: t

Formula:

• v = 20 – 5t

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

5  <  v  <  15

5  <  20 − 5t  <  15

Start with:		5  <  20 − 5t  <  15
Subtract 20 from each:		5 − 20  <  20 − 5t − 20  <  15 − 20
Simplify:		−15  < −5t  <  −5
Divide each by 5:		−3  < −t  <  −1
Change signs and reverse inequalities:		3  >  t  >  1
It is neater to show the smaller number first, so:
Swap over (for a neater answer):		1  <  t  <  3

So the velocity is between 5 m/s and 15 m/s between 1 second and 3 seconds 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:

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 bycompleting the square:

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 −32 = 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

Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

Sam and Alex play Tennis.
On the weekend Sam played 4 more games than Alex did, and together they played 12 games.
How many games did Alex play?

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

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

Also look for key words:

When you see		Think
add, total, sum, increase, more, combined, together		+
minus, less, difference, fewer, decreased		−
multiplied, times, of, product, factor		×
divided, quotient, per, out of, ratio		÷

Thinking Clearly

Some wording can be tricky, making you think "the wrong way around", such as:

Example: on our street there are twice as many dogs as cats. How do we write this as an equation?

• Let D = number of dogs
• Let C = number of cats

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

(2D = C is a common mistake, as the question is written "twice ... dogs ... cats")

Example: Sam has 2 books less than Alex. How do we write this as an equation?

• Let S = number of books Sam has
• Let A = number of books Alex has

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

Examples

Let's start with a really simple example so we see how it's done:

Example: A rectangular garden is 12m by 5m, what is its area?

Turn the English into Algebra:

Sketch:

Letters:

• Use w for width of rectangle: w = 12m
• Use h for height of rectangle: h = 5m

Formula for Area of a Rectangle: A = w × h

We are being asked for the Area.

Solve:

A = w × h = 12 × 5 = 60 m²

The area is 60 square meters.

Now let's try the example from the top of the page:

Example: Sam and Alex play Tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play?

Turn the English into Algebra:

Letters:

• Use S for how many games Sam played
• Use A for how many games Alex played

We know that Sam played 4 more games than Alex, so: S = A + 4

And we know that together they played 12 games: S + A = 12

We are being asked for how many games Alex played: A

Solve:

Start with:		S + A = 12
S = A + 4, so we can substitute "A + 4" for S:		(A + 4) + A = 12
Simplify:		2A + 4 = 12
Subtract 4 from both sides:		2A = 12 - 4
Simplify:		2A = 8
Divide both sides by 2:		A = 4

Which means that Alex played 4 games of tennis.

Check: Sam played 4 more games than Alex, so Sam played 8 games. Together they played 8 + 4 = 12 games. Yes!

A slightly harder example:

Example: Alex and Sam also build tables.

Together they make 10 tables in 12 days.

Alex working alone can make 10 in 30 days.

How long would it take Sam working alone to make 10 tables?

Turn the English into Algebra:

Letters:

• Use a for Alex's work rate
• Use s for Sam's work rate

12 days of Alex and Sam is 10 tables, so: 12a + 12s = 10

30 days of Alex alone is also 10 tables: 30a = 10

We are being asked how long it would take Sam to make 10 tables.

Solve:

30a = 10, so Alex's rate (tables per day) is: a = 10/30 = 1/3

Start with:		12a + 12s = 10
Put "1/3" for a:		12(1/3) + 12s = 10
Simplify:		4 + 12s = 10
Subtract 4 from both sides:		12s = 6
Divide both sides by 12:		s = 6/12
Simplify:		s = 1/2

Which means that Sam's rate is half a table a day (faster than Alex!)

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

Example: Jenna is training hard to qualify for the National Games.

She has a regular weekly routine, training for five hours a day on some days and 3 hours a day on the other days.

She trains altogether 27 hours in a seven day week. On how many days does she train for five hours?

Letters:

• The number of "5 hour" days: d
• The number of "3 hour" days: e

We know there are seven days in the week, so: d + e = 7

And she trains 27 hours in a week, with d 5 hour days and e 3 hour days: 5d + 3e = 27

We are being asked for how many days she trains for 5 hours: d

Solve:

		d + e = 7
So:		e = 7 − d
Put that in 5d + 3e = 27		5d + 3(7−d) = 27
Simplify:		5d + 21 − 3d = 27
Subtract 21 from both sides:		5d − 3d = 6
Simplify:		2d = 6
Divide both sides by 2:		d = 3

The number of "5 hour" days is 3

Check: She trains for 5 hours on 3 days a week, so she must train for 3 hours a day on the other 4 days of the week.

3 × 5 hours = 15 hours, plus 4 × 3 hours = 12 hours gives a total of 27 hours

Some examples from Geometry:

Example: A circle has an area of 12 mm², what is its radius?

Letters:

• Use A for Area: A = 12 mm²
• Use r for radius

And the formula for Area is: A = π r²

We are being asked for the radius.

Solve:

We need to rearrange the formula to find the area

Start with:		A = π r2
Swap sides:		π r2 = A
Divide both sides by π:		r2 = A / π
Take square root of both sides:		r = √(A / π)
Now we can use the formula:		r = √(12 / π)
And we get:		r = 1.954 (to 3 places)

Example: A cube has a volume of 125 mm³, what is its surface area?

Make a quick sketch:

Letters:

• Use V for Volume
• Use A for Area
• Use s for side length of cube

Formulas:

• Volume of a cube: V = s³
• Surface area of a cube: A = 6s²

We are being asked for the surface area.

Solve:

First work out s using the volume formula:

Start with:		V = s3
Swap sides:		s3 = V
Take cube root of both sides:		s = ∛(V)
And we get:		s = ∛(125) = 5

Now we can calculate surface area:

Start with:		A = 6s2
And we get:		A = 6(5)2
		A = 6 × 25 = 150 mm2

An example about Money:

Example: Joel works at the local pizza parlor. When he works overtime he earns 1¼ times the normal rate.

One week Joel worked for 40 hours at the normal rate of pay and also worked 12 hours overtime. If Joel earned $660 altogether in that week, what is his normal rate of pay?

Letters:

• Joel's normal rate of pay: $N per hour

Formulas:

• Joel works for 40 hours at $N per hour = $40N
• When Joel does overtime he earns 1¼ times the normal rate = $1.25N per hour
• Joel works for 12 hours at $1.25N per hour = $(12 × 1¼N) = $15N
• And together he earned $660, so:

$40N + $(12 × 1¼N) = $660

We are being asked for Joel's normal rate of pay $N.

Solve:

Start with		$40N + $(12 × 1¼N) = $660
Simplify:		$40N + $15N = $660
Simplify more:		$55N = $660
Divide both sides by 55:		$N = $12

So Joel’s normal rate of pay is $12 per hour

Check

Joel’s normal rate of pay is $12 per hour, so his overtime rate is 1¼ × $12 per hour = $15 per hour. So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660

More about Money, with these two examples involving Compound Interest

Example: Alex puts $2000 in the bank at an annual compound interest of 11%. How much will it be worth in 3 years?

This is the compound interest formula:

So we will use these letters:

• Present Value PV = $2,000
• Interest Rate (as a decimal): r = 0.11
• Number of Periods: n = 3
• Future Value (the value we want): FV

We are being asked for the Future Value: FV

Solve:

Start with:		FV = PV × (1+r)n
Put in what we know:		FV = $2000 × (1+0.11)3
Calculate:		FV = $2000 × 1.367631
Calculate:		FV = $2735.26 (to nearest cent)

Example: Roger deposited $1,000 into a savings account. The money earned interest compounded annually at the same rate. After nine years Roger’s deposit has grown to $1,551.33

What was the annual rate of interest for the savings account?

This is the compound interest formula:

So we will use these letters:

• Present Value PV = $1,000
• Interest Rate (the value we want): r
• Number of Periods: n = 9
• Future Value: FV = $1,551.33

We are being asked for the Interest Rate: r

Solve:

Start with:		FV = PV × (1+r)n
Put in what we know:		$1,551.33 = $1000 × (1+r)9
Swap sides:		$1000 × (1+r)9 = $1,551.33
Divide both sides by 1000:		(1+r)9 = $1,551.33 / $1,000
Simplify:		(1+r)9 = 1.55133
9th root:		1+r = 1.55133(1/9)
Calculate:		1+r = 1.05
Calculate:		r = 0.05 = 5%

So the annual rate of interest is 5%

Check: $1,000 × (1.05)⁹ = $1,000 × 1.55133 = $1,551.33

And an example of a Ratio question:

Example: At the start of the year the ratio of boys to girls in a class is 2 : 1

But now, half a year later, four boys have left the class and there are two new girls. The ratio of boys to girls is now 4 : 3

How many students are there altogether now?

Letters:

• Number of boys now: b
• Number of girls now: g

The current ratio is 4 : 3

b	=	4
g		3

Which can be rearranged to 3b = 4g

At the start of the year there was (b + 4) boys and (g − 2) girls, so the ratio was 2 : 1

b + 4	=	2
g − 2		1

Which can be rearranged to b + 4 = 2(g − 2)

We are being asked for how many students there are altogether now: b + g

Solve:

Start with:		b + 4 = 2(g − 2)
Simplify:		b + 4 = 2g − 4
Subtract 4 from both sides:		b = 2g − 8
Multiply both sides by 3 (so we get 3b):		3b = 6g − 24
Now substitute 4g for 3b:		4g = 6g − 24
Subtract 6g from both sides:		−2g = − 24
Divide both sides by −2:		g = 12

There are 12 girls!

And 3b = 4g, so b = 4g/3 = 4 × 12 / 3 = 16, so there are 16 boys

So there are now 12 girls and 16 boys in the class, making 28 students altogether.

Check

There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3
At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1

And for these we need to solve Quadratic Equations:

Example: The product of two consecutive even integers is 168. What are the integers?

Consecutive means one after the other. And they are even, so they could be 2 and 4, or 4 and 6, etc.

We will call the smaller integer n, and so the larger integer must be n+2

And we are told the product (what we get after multiplying) is 168, so we know:

n(n + 2) = 168

We are being asked for the integers

Solve:

Start with:		n(n + 2) = 168
Expand:		n2 + 2n = 168
Subtract 168 from both sides:		n2 + 2n − 168 = 0

That is a Quadratic Equation, and there are many ways to solve it. Using the Quadratic Equation Solver we get −14 and 12.

Check −14: −14(−14 + 2) = (−14)×(−12) = 168 YES

Check 12: 12(12 + 2) = 12×14 = 168 YES

So there are two solutions: -14 and -12 is one, 12 and 14 is the other.

Note: we could have also tried "guess and check":

• We could try, say, n=10: 10(12) = 120 NO (too small)
• Next we could try n=12: 12(14) = 168 YES

But unless we remember that multiplying two negatives make a positive we might overlook the other solution of (−14)×(−12).

And:

Example: You are an Architect. Your client wants a room twice as long as it is wide. They also want a 3m wide veranda along the long side.

Your client has 56 square meters of beautiful marble tiles to cover the whole area.

What should the length of the room be?

Let's first make a sketch so we get things right!:

Letters:

• the length of the room: L
• the width of the room: W
• the total Area including veranda: A

We know:

• the width of the room is half its length: W = ½L
• the total area is the (room width + 3) times the length: A = (W+3) × L = 56

We are being asked for the length of the room: L

Solve:

Start with:		(W + 3) × L = 56
Substitute W = ½L:		(½L + 3) × L = 56
Simplify:		½L2 + 3L = 56
Multiply all terms by 2:		L2 + 6L = 112
Subtract 112 from both sides:		L2 + 6L − 112 = 0

This is a quadratic equation, there are many ways to solve it, this time let's usefactoring:

Start with:		L2 + 6L − 112 = 0
Two numbers that multiply to give ac=−112, and add to give b=6 are 14 and −8:		L2 + 14L − 8L − 112 = 0
Group:		L(L +14) − 8(L + 14) = 0
Group:		(L − 8)(L + 14) = 0

And so L = 8 or -14

There are two solutions to the quadratic equation, but only one of them is possible since the length of the room cannot be negative!

So the length of the room is 8 m

Check

L = 8, so W = ½L = 4

So the area of the rectangle = (W+3) × L = 7 × 8 = 56

There we are ...

... I hope these examples will help you get the idea of how to handle word questions. Now how about some practice?