Question 1089671
your equations are:


x^2 + y^2 = 3


x - y = 2


right off the bat, the book can't be right when it says x is plus or minus 1.


assuming x is 1, then x^2 + y^2 = 3 becomes 1 + y^2 = 3 which becomes y^2 = 2 which becomes y = plus or minus sqrt(2).


assuming x is -1, you get the same value for y as y = sqrt(2) becauswe (-1)^2 is also equal to 1.


however:


x - y = 2 becomes:


1 - sqrt(2) = 2 which is false, so x = plus or minus 1 can't be the solution because it doesn't solve both equations simultaneosuly, which i assume is what the problem wants you to solve.


your solution of x = sqrt(14)/2 doesn't look right either.


i solved it as follows:


start with:


x^2 + y^2 = 3
x - y = 2


these are 2 equations that need to be solved simultaneously.


solve for y in the second equation to get y = x - 2


replace y with x-2 in the first equation to get x^2 + (x-2)^2 = 3


simplify to get x^2 + x^2 - 4x + 4 = 3


combine like terms to get 2x^2 - 4x + 4 = 3


subtract 3 from both sides of the equation to get:


2x^2 - 4x + 1 = 0


factor this quadratic to get:


x = 1.7071067811865 or x = 0.29289321881345


i used an online quaderatic solver at <a href = "https://www.mathsisfun.com/quadratic-equation-solver.html" target = "-blank">https://www.mathsisfun.com/quadratic-equation-solver.html</a>


i also solved it manually using the quadratic formula and got:


x = 1 + sqrt(2)/2 or x = 1 - sqrt(2)/2


this resulted in:


x = 1.707106781 or x = .2928932188


the results are the same with the exception of the number of decimal digits displayed.