maths question dunno how to do?

Let's denote the approximate change in x by q. The quantity q is assumed small (in absolute value), so we'll be able to drop q^2, q^3, and so on. Now, let's see what the problem says. It says two things:

(1) For some value of x (denote this value by x0), we have: 6 = y(x0) = x0^2 - 3*x0 + 2.

(2) By changing x0 to x0+q, the result is to change y from 6 to 6+p. So: 6 + p = (x0 + q)^2 - 3*(x0 + q) + 2

Expanding the latter square, we get (x0 + q)^2 = x0^2 + 2*x0*q + q^2. However, we agreed to drop q^2 and higher powers of q, since q is small in absolute value. So, droppin them and collecting all the terms containing q separately, we get:

6 + p = x0^2 - 3*x0 + 2 + [2*x0*q - 3*q] = 6 + [2*x0*q - 3*q]

Subtracting 6 from both sides, we get:

p = 2*x0*q - 3*q = (2*x0 - 3) * q

You can now divide both sides by (2*x0 - 3) , and that will give you q.

In the system of equations y=-3x-2 and y=x-2, the solutions are x = 0, y = -2...

I'm not going to do your homework for you, but I'll give you a couple of hints.

You need to find the first derivative. The 6 is meaningless, only the p is important.

Good catch liberal_dude, I missed the "approximate" reference.

I hate maths:|