Search over 10,000 free study notes

First begin with rearranging the second equation to put it in terms of Y. Then substitute by putting the second equation (y=3x+13) into equation one as Y^2. Expand (3x+13)^2 and add like terms to form 10x...

Factorising gives: y = (x+2)(x+1) = 0. y only equals 0 when (x+2)=0 or (x+1)=0. Therefore x=-2 or x=-1

4^(n)-1 factors to (2^(n)-1)(2^(n)+1) by difference of two squares by definition of a prime (only has factors 1 and itself (given that itself isn't 1))So Exactly one of the brackets should = 1 for 4^(n)-1...