# X2 + y2 = 8 x - y = 0? Solutions

1. X2 + y2 = 8  (equation 1) 2. X - y = 0  (equation 2)  x = 0 + y  (express x in terms of y from equation 2) x = y  (collect terms)  x2 + x2 = 8  (substitute value of y into equation 1) x4 = 8  (apply exponent law) (x4)(1/4) = 8(1/4)  (eliminate the exponent) x = 2  (simplify)  We know that x = y. This concludes that y = 2.  Check:  x2 + y2 = 8  (eqaution 1) 22 + 22 = 8  (substitue values) 4 + 4 = 8  (simplify exponents) 8 = 8  (collect terms; correct!)  x - y = 0  (equation 2) 2 - 2 = 0  (substitue values) 0 = 0  (collect terms; correct!)  Or perhaps you intended:  1. 2x + 2y = 8  (equation 1) 2. X - y = 0  (equation 2)  x = 0 + y  (express x in terms of y from equation 2) x = y  (collect terms)  2x + 2x = 8  (substitute value of y) 4x = 8  (collect terms) 4x/4 = 8/4  (divide by 4) x = 2  We know that x = y, so y = 2.  Check:  2x + 2y = 8  (equation 1) 2(2) + 2(2) = 8  (substitute values) 4 + 4 = 8  (simplify) 8 = 8  (collect terms; correct!)  x - y = 0  (equation 2) 2 - 2 = 0  (substitue values) 0 = 0  (collect terms; correct!)
thanked the writer.
Oddman commented
Your "apply the exponent law" step is in error. One only adds term exponents when multiplying terms, not when adding terms. In any event 8^(1/4) is not 2. 8^(1/3) = 2.
The second equation means
x = y
So the first equation can be rewritten as
x^2 + x^2 = 8
x^2 = 4    (divide by 2)
x = ±2

The solutions are (x, y) = (-2, -2) or (2, 2).
thanked the writer.