The question asks me to determine the intersection 3x-y=11 and 3x2-y2 = 47 This is the given answer.

y=3x-11

3x2-(3x-11)2=47

3x2 -9x2 +66x-121-47=0

-6x2+ 66x - 168= 0

x2 -11x +28 = 0

..... and so on

The problem I'm having is in line 3. I'm not sure where the '66x' came from?

Originally Posted by Urpes
3x2-(3x-11)2=47
$(3x-11)^2=9x^2-(2 \cdot 11 \cdot 3)x+121=9x^2-66x+121$.

Originally Posted by Phantasma
$(3x-11)^2=9x^2-(2 \cdot 11 \cdot 3)x+121=9x^2-66x+121$.
Is there anyway you can explain this step by step.

I can see that 3x squared = 9x2 and I see that -11 squared = 121, however I don't know where the (2 x 11 x 3)x came from :S

Also in the given answer for the question it says that instead of 121 it's -121.. im so confused.

$3x^2-(3x-11)^2=47 \\ 3x^2-(3x-11)(3x-11)=47 \\ 3x^2-(9x^2 - 2(3\cdot 11)x+121)=47 \, (\text{Multiplying through the binomial expressions using the distributive property}) \\ 3x^2 -9x^2 + 66x - 121 = 47 \, (\text{Multiplying the expression in parenthesis by -1, using the distributive property again}) \\ (3-9)x^2+66x-(121+47)=-6x^2+66x-168=0 \, (\text{Combining like terms using associative property}) \\ \footnotesize{\left(x-\left(\frac{-66+\sqrt{66^2-4(-6)(-168)}}{2(-6)}\right)\right)\left(x-\left(\frac{-66-\sqrt{66^2-4(-6)(-168)}}{2(-6)}\right)\right)=0} \, (\text{Factoring}).$

Is this kind of what you wanted?

Yes thank you very much, I see how it was done now, however I highly doubt I will ever be able to do it myself on a test. I had to stare at this for a while and I'm still having trouble understanding it completely.

Thank you though, I've definitely come as close to understanding it as I probably ever will.