How could I use simultaneous equations to solve this?

Because both equations equal y it follows that
x^2+2x-3 = 2x+1

take 2x from each side and you get
x^2 -3 = 1

add 3 to each side and
x^2 = 4
so x = 2

putting 2 for x into the equation
y=2x+1 we get
y=4+1
y=5

Test this in the other equation
y=x^2+2x-3
y=4 + 4 -3

and so yes y does equal 5

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Since x^2 =4
x can also be -2
and if that is the value of x then the value of y is:

y=2x+1
y= -4+1
y= -3

y = x^2 + 2x - 3

--------

y = 2x + 1
x^2 + 2x - 3 = 2x + 1
x^2 + 2x - 2x - 3 - 1 = 0
x^2 - 4 = 0
(x + 2)(x - 2) = 0

x + 2 = 0
x = -2

x - 2 = 0
x = 2

x = -2 , 2

--------

y = x^2 + 2x - 3
y = (-2)^2 + 2(-2) - 3
y = 4 + (-4) - 3
y = 4 - 4 - 3
y = -3

y = x^2 + 2x - 3
y = (2)^2 + 2(2) - 3
y = 4 + 4 - 3
y = 5

y = -3 , 5

The two expressions for y must be equal

x^2 + 2x - 3 = 2x +1

x^2 - 4 = 0

(x + 2)(x - 2) = 0

x = -2 and x = + 2 are the solutions