SOLUTION: x^2 + y^2 = 8 x - y = 2 My answer is (1 + sqrt5, -1 + sqrt5) and (1 - sqrt5, -1 - sqrt5) The textbook's answer is exactly the same, except that instead of sqrt5 it's sqrt3

Algebra ->  Expressions-with-variables -> SOLUTION: x^2 + y^2 = 8 x - y = 2 My answer is (1 + sqrt5, -1 + sqrt5) and (1 - sqrt5, -1 - sqrt5) The textbook's answer is exactly the same, except that instead of sqrt5 it's sqrt3       Log On


   

Question 1095507: x^2 + y^2 = 8
x - y = 2
My answer is (1 + sqrt5, -1 + sqrt5) and (1 - sqrt5, -1 - sqrt5)
The textbook's answer is exactly the same, except that instead of sqrt5 it's sqrt3
How did I get sqrt5 instead of sqrt3? I know it must be some silly mistake, but I've looked and looked and my brain's about fried.
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 + y^2 = 8
x - y = 2
My answer is (1 + sqrt5, -1 + sqrt5) and (1 - sqrt5, -1 - sqrt5)
The textbook's answer is exactly the same, except that instead of sqrt5 it's sqrt3
How did I get sqrt5 instead of sqrt3? I know it must be some silly mistake, but I've looked and looked and my brain's about fried.
=====================
No way to know how you got that, unless you show how you got it.

Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!
.
1.  First, you can easily check your solution.

    %281%2Bsqrt%285%29%29%5E2 + %28-1%2Bsqrt%285%29%29%5E2 = 

    1+%2B+2%2Asqrt%285%29+%2B+5 + 1+-+2%2Asqrt%285%29+%2B+5 = 1 + 1 + 5 + 5 = 12.


        Is it equal to 8 ? - NO.  
    
        What does it mean ?  - It means that your solution is incorrect.



2.  Very straightforward way is THIS: From the second equation express x = 2 + y

    and substitute it into first equation.  You will get

    %282+%2B+y%29%5E2 + y%5E2 = 8.

    It is your quadratic equation for single unknown "y".

    Simplify and solve it.


To see many other similar solved problems (your samples), look into the lessons
    - Solving the system of algebraic equations of degree 2 and degree 1
    - Solving the system of algebraic equations of degree 2
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Systems of equations that are not linear".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.