Basic Algebra

Overview


Every one does alge­bra every day. How do I fairly share 12 sweets between 4 peo­ple? If a recipe calls for 2oo grams of flour to make a dish for 4 peo­ple how many grams will I need to make the the dish for 2 peo­ple? I’ve got £30 mil­lion left and 6 play­ers to pick to com­plete my fan­tasy foot­ball team, how much, on aver­age, can I spend on each player?

We solve such prob­lems instinc­tively but there is some­thing about the nota­tion of alge­bra (x, y, brack­ets etc) that is intim­i­dat­ing. “x” and “y” or sim­i­lar terms are just short­hand for some unknown vari­able. “x” could refer to “num­ber of sweets” or “grams of flour” or “aver­age spend per player in £”. It’s just quicker to write for­mu­las (and set ques­tions) using terms such as “x” and “y”.

Look at this ques­tion to see how an every­day, com­mon sense prob­lem can be solved using words and by sum­maris­ing those words into nota­tions used in algebra.

Ques­tion

The Old Traf­ford grounds­man is try­ing to work out how long it will take to re-turf the pitch. The pitch mea­sures 100 meters x 60 meters (6,000m²). The grounds­men and his staff can lay 60m² of turf per hour.

a) Write this prob­lem as a for­mula using the terms; “Time taken to re-turf the whole pitch”, “Total pitch size” and “Turf laid per hour”

b) Re-write this prob­lem as a for­mula sub­sti­tut­ing “x” for “Time taken to re-turf the whole pitch” and the appro­pri­ate num­bers for “Total pitch size” and “Turf laid per hour”.

c) Cal­cu­late the num­ber of hours to re-turf the pitch.

Answer

a) Time taken to re-turf the whole pitch  = Total pitch size/Turf laid per hour.

b) x = 6,000m²/60m²

c) 100 hours

As usual here are some more ques­tions fol­lowed by my approach (which hope­fully explains meth­ods and tech­niques used) and answers.

Alge­bra Questions.

1. Write an expres­sion for the area of a rec­tan­gle with a length of 6x + 3 and a width of x.

Expand this expression.

2. Show that 6(x+2) + 8(x+1)  ≡   2(7x +10)

3. A reg­u­lar octa­gon has sides of length 9x + 3 metres.

Write an expres­sion for the perime­ter of the octagon.

If the perime­ter of the octa­gon is 240 metres, what is the value of x?

How long is each side of the octagon?

4. Fac­torise  12pr³ + 18p²r² + 6p²r

5. Show that (4x + 4) (2x — 5) ≡  4(2x² — 3x — 5)

6. Show that (x + y)² — (x — y)² ≡ 4xy

Use this knowl­edge to cal­cu­late 69² — 51²

Alge­bra Approach.

1. & 3. Where pos­si­ble you should use brack­ets. Remem­ber that W(Y +Z) is equal to W x (Y + Z) and that each ele­ment in the bracket needs to mul­ti­plied by the term out­side the bracket so W(Y+Z) = W x (Y + Z) = (W x Y) + (W x Z). Expand­ing an impres­sion refers to this process of mul­ti­ply­ing out the brackets.

2. The sign “≡” means iden­ti­cal to. There are often ques­tions about show­ing that one expres­sion is iden­ti­cal to another expres­sion. Just expand each expres­sion step by step to demon­strate they are identical.

4. Fac­toris­ing is the oppo­site of expand­ing an impres­sion. You find com­mon fac­tors to place out­side brack­ets and place the remain­ing terms in brack­ets. For exam­ple 2x² + 4x becomes 2x(x + 2). You can do this in stages:

2x² + 4x. Remov­ing the first com­mon fac­tor, x gives:- x(2x + 4)

x(2x + 4) now remove the final com­mon fac­tor, 2 gives:-  2x(x +2)

It is best to check your answer by expand­ing it, in this case:

2x(x + 2) =2x² + 4x.

It can help to write an expres­sion with the pow­ers mul­ti­plied out to help you under­stand the com­mon fac­tors so:

12pr³ + 18p²r² + 6p²r  = (12 x p x r x r x r) + (18 x p x p x r) + (6 x p x p x r)

Then write down the com­mon fac­tors in this case :- 6 (relates to 12, 18 & 6), p & r. These fac­tors then go out­side the bracket and the remain­ing terms are placed in the bracket.

5. Expand each expres­sion. This type of ques­tion is good as you know that if you can show that the expres­sions are iden­ti­cal you have got it right!

6. When you are squar­ing a bracket use this method (a + b) (a + b) =

a² + ab + ba + b² = a² +2ab + b²

Remem­ber that mul­ti­ply­ing a neg­a­tive by a neg­a­tive gives a pos­i­tive so (a — b) (a — b) =

a² — ab + –ab + b² = a²  –2ab + b²

This is another type of ques­tion that gives instant encour­age­ment and is sat­is­fy­ing to solve.

Alge­bra Answers.

1. 3x(x +1). Expanded =  6x² + 3x

2. 6(x+2) + 8(x +1) = 6x + 12 + 8x + 8 = 14x + 20

2(7x + 10) = 14x + 20

3. Octa­gon has 8 sides so if each side is 9x + 3 then the perime­ter is:

8(9x + 3)

If the perime­ter is 240 metres then:-

8(9x + 3) = 240

9x + 3 = 30

9x = 27

x = 3

Each side = (9 x 3) + 3 = 30 metres

4. High­est com­mon fac­tors = 6, p and r.

Answer = 6pr(2r² + 3pr + p)

5. (4x + 4) (2x — 5) = 8x² — 20x + 8x –20 = 8x² — 12x — 20

4(2x² — 3x — 5) = 8x² — 12x — 20

 

6. (x + y)² = (x + y) (x + y) = x² +xy + yx + y² = x² + 2xy + y²

(x — y)² = (x — y) (x — y) = x²  –xy — yx + y² = x² — 2xy + y²

There­fore (x + y)² — (x — y)² = (x² + 2xy + y²) — (x² — 2xy + y²) = 4xy

and (x + y)² — (x — y)² ≡ 4xy

69² — 51² = (60 + 9)²  — (60 — 9)²

Since we know (x + y)² — (x — y)² ≡  4xy it fol­lows that

(60 + 9)²  — (60 — 9)² = 4 x 60 x 9 = 4 x 540 = 2,160

and 69² — 51² = 2,160

This entry was posted in 12. Basic Algebra and tagged , , , , , , . Bookmark the permalink.

One Response to Basic Algebra

  1. luul omer says:

    i’m doing GCSE

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