Can you please explain number 5 and 8 on the working with advanced system of equations.

Sure! (And sorry! I see you submitted this a few days ago but for some reason I’m only seeing it now.)

Since #5 is a no-calc question, let’s get right to solving for x by substituting.

2(x-4)^2-2=(x-3)^2\\2(x^2-8x+16)-2=x^2-6x+9\\2x^2-16x+30=x^2-6x+9\\x^2-10x+21=0\\(x-7)(x-3)=0

So we know the equations intersect at x=7 and x=3. Let’s plug those x-values into one of the equations to get y-values.

y=(3-3)^2=0\\y=(7-3)^2=16

The two points we need, then, are (3, 0) and (7, 16). Get the slope:

\text{slope}=\dfrac{16-0}{7-3}=4

For #8, you can use your calculator to play around here, plugging in values for a and b, but the concept being tested is that you have two parabolas with vertices at (5, 5) and (5, 6), and the coefficients a (for the upper) and b (for the lower) tell you how wide those parabolas will be. Remember, the greater a positive coefficient, the thinner the parabola will be.

If the lower parabola is skinnier than the upper one (a<b), they’ll definitely intersect. However, if they’re both exactly as wide (a=b), they’ll never intersect, and if the lower one is wider than the upper one (a>b), they’ll never intersect.

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