Help! 
Maths (IGCSE)
May June 2000
Q1.
(a) (i) (ii) (iii)
(b)
How do you find them? 
SORRY FO THE DELAY !! .. my computer shut down unexpectedly and had to type ALL of this again !! =@
also dunno if the diag might help much..
For these types of sums always remember “y-intercept” (
c )
y = mx +
cthis is a common equation for all equations of graphs ..
y-intercept means where the equations ‘cuts’ the y-axis
in this one y= x + 2 .. it cuts at 2 .. u got 1.a)
so the coordinates of P would be (0,2)for ii) if u notice Q is at y=0
so substitute 0 in 3x + 4y = 22 to find x
tht would be
3x + 4 x 0 = 22
3x = 22
x = 7.3
so coordinates of Q = (7.3 , 0)iii) for this R is the intersection .. so tht means at tht point BOTH the equations should be equal ..
so y = x+2 (eqn1) would be equal to 3x + 4y = 22 (eqn2)
so lets find the y for R ..
rearrange 3x + 4y = 22 to
y = 22 – 3x
---------
4
As eqn1 = eqn2
22 – 3x
---------- = x + 2
4
22 – 3x = 4x + 8
7x =14
X = 2
Now put “x=2” in any of the eqn ..
U’ll get y =4
so coordinates for R = (2 , 4)b) is a direct question .. u hav to state some inequalities that lie in the region OPRQ.
y >= 0
y =< x + 2
3x + 4y =< 22
Define this region .. for this u need understand the concept .. two eqns are already given .. the region is ‘less than’ both of them .. third would be
y >= 0 there is no negative quadrant
..
