Answer:
(-1, -3) and (3, 5)
Explanation:
I find a graphing calculator to be a very useful tool for finding solutions to problems like this. The attachment shows the points of intersection to be (-1, -3) and (3, 5).
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You can find these points algebraically by substituting for y in either equation using the expression provided by the other equation.
x^2 -4 = y = 2x -1
x^2 -2x +1 = 4 . . . . . . . add 5-2x to make a perfect square trinomial
(x -1)^2 = 4 . . . . . . . . . show as a square
x -1 = ±√4 = ±2 . . . . . take the square root
x = 1 ±2 = {-1, 3} . . . . add 1, show the separate solutions
y = 2x -1 = 2{-1, 3} -1 = {-2, 6} -1 = {-3, 5} . . . . find the corresponding y-values
The points of intersection are (-1, -3) and (3, 5).