Alegbra class

Last night's homework question 2 had a simultaneous equation - one linear one quadratic - type question. Notethat it was xy and not x², but it has to be handled the same way.
x + 3 = 2y
xy – 7y + 8 = 0 <-this is the quadratic (can be x² or xy)
x = 2y -3 [rearrange linear equation]
(2y-3)(y) – 7y + 8 = 0 [sub in new value for x]
2y² – 3y -7y +8 = 0
2y² – 10y + 8 = 0 [always simplify equation if you can]
y² -5y + 4 = 0 [factor -4 x -1 = 4 and -4 -1 = -5]
(y-1)(y-4) = 0
So y = 1,4
If y= 1 then
x = 2(1) – 3, so x = -1 values: x = -1, y = 1
if y= 4 then
x = 2(4) – 3, so x = 5 values: x = 5, y = 4
And always check:
(-1)(1) – 7(1) + 8 = 0
(5)(4) – 7(4) + 8 = 0

Part c of this question works out as follows:
(i) √x + 1 [Turn them into same type of fraction ... LCM of 1 and √x is √x]
1 √x
= √x√x + 1 [multiply above and below leftmost fraction by √x]
√x √x
= x + 1 [don't forget √x√x = x]
√x

ii) ( 2√x ) • (x + 1 ) [cancel (x + 1) with (1+x) and cancel √x with 2√x ]
(1+x ) ( √x )
= 2

iii) 2 = x -3
x = 5

Tomorrow's double class will be Perimeter area + volume (1st period) and Linear Programming (2nd period) - Homework for tomorrow is 2006 Paper 2 - q1 and 11.