Using algebra to solve problems

We focussed today on the type of questions where an unknown is parcelled up in two different ways and a relationship between them is given.

There is no doubt about it - these problems can be tricky. Partly because you have to parse through a lot of text, and partly because the equation that you end up with can be awkward.

Some good examples are in the 2006, 2005 and 2004 papers.

In the 2006 q, you make 2 expressions
540/x gives us the number of days the first lot of hay will last.
300/x+1 gives us the number of days the second lot will last
The sum of these two expressions is 90 days.

When you build your equation and simplify it you get

3x² -25x - 18 = 0

After you have factored, don't forget to discard the negative root and write your answer as x = 9 bales of hay and to work out the last part of the problem ... how many days it took to complete the first set of 540 bales. In other words, evaluate the first expression with x = 9 and write down the answer as 60 days.