Solve for x and y : x^2-y^2=4; x+y=2
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We notice that the 1st equation is a difference of two squares that returns the product:
x^2 - y^2 = (x-y)(x+y) = 4
We also notice that the second factor of the product represents the 2nd equation of the system.
(x-y)(x+y) = 4 <=> (x-y)*2 = 4
We'll divide by 2:
x - y = 2
We'll add this equation to the 2nd:
x - y + x + y = 2 + 2
2x = 4
x = 2
x + y = 2 => y = 0
The solution of the system is represented by the pair (2 ; 0).
From the rule of factoring, we know
the first equation becomes:
we know that x+y =2
substitute into the equation
and we know x+y=2
add the two functions together
substitute x back into the original function
So the solution of the original function pair is x=2, y=0
x^2 - y^2 = 4
x + y = 2
This system can also be solved graphically. First, set each equation to the form y=
x^2 - y^2 = 4 y = sqrt(x^2 - 4)
x + y = 2 y = 2 - x
Here is the graph:
The point of intersection of the two graphs is (2, 0).
Therefore, x = 2 and y = 0.
If you do not have a graphing calculator, refer to the attached website. It will graph equations and calculate the point of intersection.
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