# Solve for x and y : x^2-y^2=4; x+y=2

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### 3 Answers

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.

x^2-y^2=4

From the rule of factoring, we know

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

the first equation becomes:

(x+y)(x-y)=4

we know that x+y =2

substitute into the equation

2*(x-y)=4

x-y=2

and we know x+y=2

add the two functions together

2x=4

x=2

substitute x back into the original function

2^2-y^2=4

4-y^2=4

y^2=0

y=0

**So the solution of the original function pair is x=2, y=0**

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).**