Expert: Josh - 3/2/2010

Question
1.solve simultaneously:
2y-x=3 and y^2-2x^2-x=1

2.determine the value of a in y=a^x which passes through the point (2;1/9)


3.a triangle: P(a;4)R(-3;-8)Q(3;-2)
PQ=QR
Find the x co-ordinate of P and assume that P is in the second quadrant

Answer
Hi Zainy,

Q1. Let's make y the subject in the first equation.
   2y-x=3 becomes y=(x+3)/2. Put this into the second equation.
   y^2-2*x^2-x=1 becomes
   (x+3)^2/4 -2*x^2 -x = 1  ...multiply both sides by 4
   (x+3)^2 -8*x^2 -4x = 4  ...expand LHS
   x^2+6x+9-8*x^2 -4x = 4
   7*x^2 -2x -5 = 0
   (7x+5)(x-1) = 0
   x=-5/7, or x=1
   Back-substitute these x values in y=(x+3)/2 to get corresponding y values.

Q2. The inverse function which "undo" "a" raised to the power of x is the logarithm base "a" function. Given y=a^x, log_a(y)=x .....[#1]

Change of logarithmic base rule:
log_a(y)=log_10(y)/log_10(a). You have log_10 on your calculation, so you can evaluate this. In fact, you can use any base you like on the RHS of the equation, you can also compute log_a(y) as ln(y)/ln(a), where "ln" is the natural logarithm function.

[#1] becomes ln(y)/ln(a)=x => ln(a)=ln(y)/x provided x is not zero.
First, plug in x=2 and y=1/9 to find ln(a).
i.e., ln(a)=ln(1/9)/2, then take the exponential function to "undo" the natural log.
i.e., a = exp(ln(1/9)/2).
Please check that I have not made silly mistakes.

Q3. Strategy: Write down distance equation between two points for line segment PQ and QR.

In fact, we will used the square distance to avoid complication with the square root.
From the Pythagoras theorem, distance = sqrt[(y2-y1)^2+(x2-x1)^2]......
So, the square distance, |PQ|^2 = (4--2)^2+(a-3)^2 = 36+(a-3)^2.  ...[#2]
Likewise, |QR|^2 = (-2+8)^2+(3+3)^2 = 72  ...[#3]

Since |PQ| = |QR|, it must also be true that |PQ|^2 = |QR|^2.
Equating the squared distances [#2] and [#3], 72=36+(a-3)^2
=> 36=(a-3)^2
=> square_root_of(36) = a-3
=> 6 = a-3
=> a=9.

Hence, P=(9,4).

Again, I haven't double checked to verify the solutions.