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

samhouston | Middle School Teacher | (Level 1) Associate Educator

Posted on

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.

shaznl1 | High School Teacher | (Level 1) Salutatorian

Posted on

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

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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).
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}