Unit 5 sec 4.3 using algebra

Unit 5 sec 4.3 using algebra

20 January 2016

22:07

To do this, you carry out the trick starting with a letter (n, say), which represents any number at all. After each step you simplify the resulting expression.

Notice that it was important to include the brackets when the expression 2n + 7 was doubled in the above calculation. Doubling 2n + 7 does not give 2 × 2n + 7 = 4n + 7; the correct calculation is

2(2n + 7) = 4n + 14.

Example 16 Finding and simplifying a formula

Each week Arthur works for at least 37 hours. He’s paid £15 per hour for

the first 37 hours, and £25 per hour for each additional hour. Find a

formula for P, where £P is Arthur’s pay for the week if he works for

n hours.

Solution

Arthur’s pay in £ for the first 37 hours is

37 × 15 = 555.

The number of additional hours that Arthur works is n − 37, so his pay

in £ for these additional hours is

(n − 37) × 25 = 25(n − 37).

The question states that Arthur works for at least 37 hours, so there’s

no need to worry about n − 37 going negative.

So a formula for Arthur’s pay is

P = 555 + 25(n − 37).

The formula can be simplified by multiplying out the brackets:

P = 555 + 25n − 925

= 25n − 370.

So the simplified formula is

P = 25n − 370.

Example 17 Proving a property of numbers

Prove that the sum of any three consecutive integers is divisible by 3.

Solution

Represent the first of the three integers by n. Then the other two integers

are n + 1 and n + 2. So their sum is

n + (n + 1) + (n + 2) = n + n + 1 + n + 2

Dividing by 3 gives

3n + 3    = 3n   + 3 = n + 1.

   3            3       3

Now n + 1 is an integer, because n is an integer. So dividing the sum of

the three numbers by 3 gives an integer. That is, the sum is divisible by 3.