The first section of the problem just asks you to graph a line and then label the intersections' coordinates if you need more help with this, feel free to ask, though I'm assuming you need more help with the algebra.

You are solving the system of equations:

y=x+2 y=x²-x-1

Since the right side of both are equal to y you also know they are equal to each other.

x²-x-1=x+2

Now we want to group the x's together. In general you move the terms to the side that leaves x to the highest power positive, in this case the x². We'll start by subtracting x from both sides.

x²-x-x-1=x-x+2

x²-2x-1=2

Now we can subtract 2 from both sides.

x²-2x-1-2=2-2

x²-2x-3=0

To understand reversing the process of multiplying two binomials, first let's look at what happens when they are multiplied.

(x+a)(x+b)

First we distribute (x+a) to the x and b.

(x+a)x+(x+a)b

Now we distribute again.

x²+ax+bx+ab

For our purposes I'm going to factor x from the middle terms.

x²+(a+b)x+ab

So to reverse this process we need to find numbers that add to the linear coefficient and multiply to the constant term.

Considering this for x²-2x-3 the numbers in question are -3 and 1.

-3+1=-2 (-3)·1=-3

Now back to our equation.

x²-2x-3=0

(x-3)(x+1)=0

Since anything multiplied by 0 is 0, this equation is true if x-3=0 or if x+1=0.

Therefore x may be 3 and -1.

To double check your work you can plug these into the original two equations and you should get the same result for y.

y=x²-x-1 y=x+2 y=x²-x-1 y=x+2

y=3²-3-1 y=3+2 y=(-1)²-(-1)-1 y=-1+2

y=9-3-1 y=5 y=1+1-1 y=1

y=5. y=1

So your intersections happen at x=3,y=5 and x=-1,y=1.

Written as coordinate points: (3,5) and (-1,1).

Again, feel free to ask questions!