MontrealLady wrote:
If #p# = ap3+ bp – 1 where a and b are constants, and #5# = 3, what is the value of #5#?
The answer I found was (5) and here is how I solved the question.
for #5# = 3, here is the equation 125a – 5b – 1 = 3
For #5#, the equation is 125a + 5b – 1 = X
then to find X, I substract the first equation from the second one
125a – 5b – 1 = 3

125a + 5b – 1 = X
that gives 2 = 3x, so x = 5
Your approach is much better than that of MGMAT's. Though there are two problems:
When you subtract equations you'll get: \(250a10b=3x\) (so you subtracted incorrectly) and this doesn't work.
What you should have done is to add equations instead of subtracting: \(125a5b1+125a+5b1=3+x\) > \(2=3+x\) > \(x=5\).
Hope it helps.
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