Algebra: Real World Applications andProblems

Algebra is boring. Right? Hopefully not. Algebra has no applications in thereal world. Wrong. Absolutely wrong. I hope to show this in the followingdocument. We’ll cover topics like forming and solving equations and showwhere we might use this outside of the classroom. We’ll move onto quadraticsand spend a little while looking at applications of quadratics and parabolas.On the way we will try to cover extracting common factors along with thedifference of two squares and show why these might be useful outside theclassroom.

Solving Equations

We begin with the basics, that is solving simple equations. The most impor-tant thing to remember when solving equations is that whatever you do toone side you need to also do to the other. We will use letters x, y and z todenote the things that we want to find out.

Take an example. Let x be the cost of a meal in a particular restaurant.Bob pays for two of these meals and the cost is £30. How much is each meal?

Solution: Well we construct an equation that represents the bill.

2x = 30 ( 2 meals x costs £ 30)

÷2 ÷2

x =30

2= 15

So the meal costs £15.

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Too easy right. I didn’t need algebra for that you’re saying. Well ok maybenot but you can use it to help you out. How about this one?

Bob goes to a different restaurant and orders the same two meals, but withdrinks this time. The drinks cost £4 (cheap I know) and the total bill comesto £28. If y represents the cost of the meal, form an equation for y and solveit to find y.

Solution: Again we construct an equation that represents the bill.

2x+ 4 = 28 ( 2 meals x + £ 4 drinks = £ 28)

−4 −4

2x = 24

÷2 ÷2

x =24

2= 12

So the meal costs £12.Still too straight forward or getting more complicated? Final example – Bobgoes to a third restaurant and his total bill is identical to another customer’s,Cindy. Bob’s order is two identical main courses, a £2 drink and a £4.50pudding with his meal whilst Cindy just has one main course (the same oneas Bob) and £16.50 worth of drinks on her bill. If z represents the cost ofthe main meal both Bob and Cindy order, form an equation to solve for zand solve it.

Solution:

2x+ 2 + 4.50 = x+ 16.50

2x+ 6.50 = x+ 16.50

−6.5 −6.5

2x = x+ 10

−x −xx = 10

So in this final restaurant the main meal costs £10.

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Try these questions:

i) Customers at a bar want to compare the price of a particular cocktail.Wendy got a £1 reduction on each of the 9 cocktails she ordered buther bill was the same as Jane’s who bought only one of the cocktailsbut also got £15 of other drinks.

(a) If we let x be the cost of the cocktail, form an equation for x.

(b) Solve the equation to find x.

ii) A supplier offers a discount on a product bought in bulk. If you buy240 of them you get £20 off each item and this reduces the total costto the same as buying 200 at their full price.

(a) If we let x be the cost of the product, form an equation for x.

(b) Solve the equation to find x.

iii) Two competing firms offer a particular item at a reduced rate. Thereductions from each firm are different. If you buy 360 of them fromfirm A, you get a £4 reduction off each item. For the same total pricefirm B gives you 320 of the same item but with a discount of £3 peritem.

(a) If we let c be the cost of the item, form an equation for c.

(b) Solve the equation to find c.

iv) Can you think of a question similar to those above that could result inthe equation

23(x− 1) = 5(x+ 3)?

Solve the equation for x.

v) Produce a question that might result in the equation

14(s− 7) + 2(s+ 21) = 0

and solve the equation to find s.

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Ok now I’ve got a real brain teaser for you. Bob would like you to work outhis age. He gives you the following clue:

One third of my age 8 years ago is equal to one fifth of how oldI will be 5 years time. How old am I?

How can we find his age? Using logic and algebra of course.

Solution: Call his age x, then his age 8 years ago is x − 8 and his age in5 years time is x+ 5. This gives us

x− 8

3=

x+ 5

5×3 ×3

x− 8 =3(x+ 5)

5×5 ×5

5(x− 8) = 3(x+ 5)

5x− 40 = 3x+ 15

+40 +40

5x = 3x+ 55

−3x −3x

2x = 55

÷2 ÷2

x =55

2= 27.5

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Here are some more problems for you to solve

i) One sixth of my age three years from now is the same as one seventhof my age four years from now or if we let x be my age then

x+ 3

6=x+ 4

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Find x.

Similar word problems give rise to these equations. Can you solvethem?

ii)x+ 11

3=x+ 14

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iii)x+ 8

2=x+ 1

3

iv)x− 9

3=x+ 4

4

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Some more problems (optional).

i) Using the equation V = Al (volume = area × length) find l if A = 7cm2

and V = 24.5cm3.

ii) Using the equation V = IR (voltage = current × resistance) find thecurrent I if the voltage is 13 volts and the resistance is 39 ohms.

iii) Using the equation P = I2R (Power = current2 × resistance) find thecurrent I if the power is 200 Watts and the resistance is 8 ohms.

iv) Using the equation v = u + at (current velocity = original velocity +acceleration × time) find the time t if v = 20m/s, u = 10m/s anda = 2m/s2.

v) Using the equation V = πR2h (which you should remember from thevolume topic), find an expression for h and then find the value of h ifV = 62m3 and R = 3m.

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Multiplying Brackets

Let us move on. Farmer Giles has a field that is 20m wide and was originally70m long. He increases the length of the field by x metres so that the totalarea of the field is now 3200m2. The picture below shows his field.

How much has he increased the length of his field by?

Solution: Well Area = length × breadth. In this case that gives

Area = 20(x+ 70) = 3200

20× (x+ 70) = 3200

20× x+ 20× 70 = 3200

20x+ 1400 = 3200

−1400 −1400

20x = 1800

÷20 ÷20

x =1800

20= 90

So the farmer has increased the length of his field by 90m.

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A more complicated example. The farmer’s cousin, Esmeralda, has a kitchenthat is 3m wide by 8m long. She is getting an extension and the architectwants a general equation to show the area of her new kitchen. Due to plan-ning restrictions the architect must increase the length and depth by thesame amount. The architect draws the diagram below to help him.

What is the general equation for the area of the new kitchen?

Solution:

Area = Length× Breadth

= (x+ 3)× (x+ 8)

= x× (x+ 8) + 3× (x+ 8)

= x× x+ x× 8 + 3× x3× 8

= x2 + 8x+ 3x+ 24

= x2 + 11x+ 24

So we multiply the brackets together by spliting up the first bracket andmultiplying each term into the second bracket. If you are met with 3 brackets,take two of the brackets and multiply them together as above. Then split thethird bracket and multiply each term by the answer you got from multiplyingthe first two together.

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Find the equations of the areas of these rooms

and find the area of these photo frames. (Hint: Find the area of the outsiderectangle and subtract the area of the inside rectangle.

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Now you can find the area of these shapes

and finally find the volume of this cuboid.

If you wish to do more practice (remember you get tested on the less in-teresting routine multiplying and rearranging of these brackets) you can useexercise 2 on page 50 of the textbook.

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Factorising

Now we will move on from multiplying things together to pulling them apart,little by little. We will talk about common factors and multiplying quadrat-ics. We begin with common factors.

Common Factors

Recognising common factors makes simplifying equations easier. What useis this you might say? Well you go to the supermarket to buy 4 apples, 4bananas, 4 boxes of strawberrys and 4 boxes of blueberrys to make smoothies.You want to make sure you’ve got enough money to pay for everything (andyou’ve forgotten anything that might have the ability to be a calculator beforeyou suggest that). Is it going to be easier to do

i) 4 × £apples + 4 × £bananas + 4 × £blueberries + 4 × £strawberriesor

ii) 4 × (£apples + £bananas + £blueberries + £strawberries)?

You will get the same answer no matter which you do and as you’re onlymultiplying by 4 it might not make too much of a difference, what if youwere multiplying by 23? I don’t know about you but I would probably doii). We can use common factors in more complicated equations/examples aswell, we’ll see if in all sorts of problems later including quadratic equationswhich is the next thing we’ll cover.

What does this sort of problem look like? Well something like thisi) Simplify 21x+ 14y+ 28z. Which could be taken as, simplify the sum thatgives us the cost of 21 packs of double A batteries, 14 nine volt batteries and28 packs of triple A batteries.Solution: We look for a common factor. This could be a number or a letteror both.

21x+ 14y + 28z = 7× 3x+ 7× 2y + 7× 4z

= 7(3x+ 2y + 4z)

So 7 is a factor of each part of the original equation and so we can move itoutside some brackets simplifying the expression.

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Now for a more complicated problem. Simplify this expression for the area ofthe shape and try to work out possible expressions for its length and width.

Solution: Again we look for a common factor.

2x2 + 2xy = 2× x× x+ 2× x× y= 2× x× (x+ y)

= 2x(x+ y)

This time there is a common factor of 2 × x, a letter and a number. So wecan remove both outside the brackets.

Try a selection of questions from exercise 4 on page 53 of your textbook.

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Quadratic Equations

First a bit of background which we will need this for later anyway. Aquadratic equation that looks like this

y = ax2 + bx+ c.

We need to learn how to break one up (factorise) into separate brackets.Remember that

(x+ 1)(x+ 1) = x2 + 2x+ 1,

well we will learn how to go in the opposite direction

x2 + 2x+ 1 = (x+ 1)(x+ 1) = (x+ 1)2.

Why on earth might we want to do this? Well, if you plot a quadraticequation you get a graph that is called a parabola. The following picturesare examples of parabolas in the real world

Projectile motion, suspension bridges, acceleration due to gravity all usequadratic equations. Why do we want to be able to factorise quadratics(split them up into two brackets)? Well say we are building a sculpture likethe St Louis Arch shown above (middle picture). We know the equationneeded to follow to make it the correct shape but we want to make sure wehave enough room on the ground for it to fit into. Solving

ax2 + bx+ c = (x− A)(x−B) = 0

by factorising the quadratic tells us where the new statue will hit the ground(at A and B).

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So thats why we might want to factorise. How do we do it? Well its allabout your times tables and a bit of trial and error. Let me show you someexamples. We’ll begin with something nice and simple.i) Factorise x2 + 4x+ 3.

Solution: We need to split x2 + 4x + 3 up into two brackets x2 + 4x + 3 =( )( ). The first term in each bracket will make the x2 and the lastterm will make the 3.

It is easier with when we have one x2 as there is a trick to speed thingsup. To make the middle term we are looking for two numbers that multiplyto give 3 and add up to give 4.

31 × 3-1 × -3

Only 1 and 3 add up to 4, so those are the numbers we’re looking for. There-fore, x2 + 4x+ 3 = (x+ 1)(x+ 3).

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ii) Let’s do something a little more complicated. Factorise x2 − 13x+ 42.

Solution: As before we’re looking to split x2 − 13x + 42 = ( )( )into 2 brackets. This time we’re looking for two numbers that multiply togive 42 but add up to give -13.

421 × 422 × 213 × 146 × 7-1 × -42-2 × -21-3 × -14-6 × -7

The pair of numbers that sums to -13 are -6 and -7, so we have.

x2 − 13x+ 42 = (x− 6)(x− 7)

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iii) One more example (slightly different). Factorise x2 − 5x− 36.

Solution: In this example we are looking for two numbers that multiplytogether to give -36 and add together to give -5.

-361 × -362 × -183 × -124 × -96 × -69 × -412 × -318 × -236 × -1

The two that add up to give -5 are 4 and -9. So we have

x2 − 5x− 36 = (x− 5)(x+ 4)

Try a selection of questions from exercise 6 on page 56 of your textbook.

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There is another type of quadratic that you need to be able to factorise.That is one where you don’t just have x2, but a multiple of it. For example:iv) Factorise 2x2 + 5x+ 2.

Solution: This type of quadratic is more complicated. We now have a mul-tiple of x2. We still need the first terms of each bracket to make 2x2 and thelast terms to make 2.

2x2

2x × xx × 2x

21 × 2-1 × -2

As all the signs in 2x2 + 5x+ 2 are positive there is no point using -1 and -2.So we have two possible combinations.

2x2 + 5x+ 2 = (2x+ 1)(x+ 2) or

= (x+ 1)(2x+ 2)

We can now use a little common sense to decide which one is the correctanswer. If we look at the question 2x2 + 35x+ 2 does not have any commonfactors but (x+ 1)(2x+ 2) has a common factor of 2 in the last bracket. Wecan’t have a common factor in the answer if there isn’t one in the question.So we must have

2x2 + 5x+ 2 = (2x+ 1)(x+ 2).

This can be confirmed by multplying the brackets out.

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v) Factorise 4x2 − 11x− 3.

Solution: This is even more complicated. We need the first terms of eachbracket to make 4x2 and the last terms to make -3.

4x2

4x × x2x × 2xx × 4x

-3-1 × 31 × -3

So we have six possible combinations.

4x2 − 11x− 3 = (4x− 1)(x+ 3) or

= (4x+ 1)(x− 3) or

= (2x− 1)(2x+ 3) or

= (2x+ 1)(2x− 3) or

= (x− 1)(4x+ 3) or

= (x+ 1)(4x− 3)

This time we can’t simplify the problem, we have 6 possibilities and we needto multiply out to find which solution fits the bill. We begin with the firstone

(4x− 1)(x+ 3) = 4x(x+ 3)− 1(x+ 3) = 4x2 + 12x− x− 3 = 4x2 + 11x− 3

which isn’t what we want. Try the second one

(4x+ 1)(x− 3) = 4x(x− 3) + 1(x− 3) = 4x2 − 12x+ x− 3 = 4x2 − 11x− 3

which is exactly what we want. Therefore the solution is

4x2 − 11x− 3 = (4x+ 1)(x− 3)

Try a selection of questions from exercise 7 on page 57 of your textbook.

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Difference of 2 Squares

This is the final part of our algebra topic and it is a special case of thepolynomial equations known as“difference of 2 squares”. If you can identifya polynomial as an example of this special case then factorisation is mucheasier.

Take an example. We want to find the shaded area in this diagram

Solution:

Shaded Area = Big Square− Little Square

= a× a− b× b= a2 − b2

But (a + b)(a − b) = a(a − b) + b(a − b) = a2 − ab + ab − b2 = a2 − b2. Sowhat we actually have is a very neat way of factorising things that look likea2 − b2.

a2 − b2 = (a+ b)(a− b)

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How about the difference between the area of these two shapes?

Solution:

Difference in Area = Big Square− Little Square

= 75x2 − 27y2

= 3(25x2 − 9y2

)removing a common factor of 3

= 3[(5x)2 − (3y)2

]= 3(5x+ 3y)(5x− 3y)

Try a selection of questions from exercise 5 on page 54 of your textbook,including some parts of question 4.

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Summary of Factorisation

There is a three step process when it comes to solving factorisation problemsin exams and homework.

i) Common factor.

ii) Difference of 2 squares.

iii) Factorise.

First you always check for a common factor it will make any other fac-torisation you have to do much easier. Secondly check if the expression is adifference of 2 squares because these are a special case. Finally, factorise.

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Solutions to Problems

Solving Equations

Page 3

i) (a) 9(x− 1) = x+ 15

(b) x = 3.

ii) (a) 240(x− 20) = 200x

(b) x = 120.

iii) (a) 360(c− 4) = 320(c− 3)

(b) c = 12.

iv) x = 199

v) x = 72

Page 5

i) x = 3

ii) x = −2

iii) x = −22

iv) x = 48

Page 6 (optional questions)

i) l = 24.57

= 3.5m

ii) I = 1339

= 13amps

iii) I2 = 2008

= 25, I =√

25 = 5amps

iv) t = 20−102

= 5secs

v) h = VπR2 , h = 62

9π= 2.193m

Page 9

1 (x+ 1)(x+ 7) = x(x+ 7) + 1(x+ 7) = x2 + 7x+ x+ 7 = x2 + 8x+ 7

2 (x+ 3)(x− 5) = x(x− 5) + 3(x− 5) = x2 − 5x+ 3x− 15 = x2 − 2x− 15

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3 (x− 2)(x− 2) = x(x− 2)− 2(x− 2) = x2 − 2x− 2x+ 4 = x2 − 4x+ 4

4 (3−x)(x+12) = 3(x+12)−x(x+12) = 3x+36−x2−12x = 36−9x−x2

Q5 14x− 12(x− 4) = 14x− 12x+ 48 = 2x+ 48

Q6 x(x+4)−(x−3)(x−1) = x2+4x− [x2−4x+3] = x2+4x−x2+4x−3 =8x− 3

Page 10

i) (x − 1)(x2 + 2x − 3) = x(x2 + 2x − 3) − 1(x2 + 2x − 3) = x3 + 2x2 −3x− x2 − 2x+ 3 = x3 + x2 − 5x+ 3

ii) (x+ 7)(3x2−2x−1) = x(3x2−2x−1) + 7(3x2−2x−1) = 3x3−2x2−x+ 21x2 − 14x− 7 = 3x3 + 19x2 − 15x− 7

iii) (x2 + 3)(x2 + 13) = x2(x2 + 13) + 3(x2 + 13) = x4 + 13x2 + 3x2 + 39 =x4 + 16x2 + 39

The volume of the cuboid is given by

(x+ 1)(x+ 4)(x− 2) = (x+ 1) [x(x− 2) + 4(x− 2)] = (x+ 1)[x2 − 2x+ 4x− 8

]= (x+ 1)

(x2 + 2x− 8

)= x(x2 + 2x− 8) + 1(x2 + 2x− 8)

= x3 + 2x2 − 8x+ x2 + 2x− 8

= x3 + 3x2 − 6x− 8

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