Question 1089671: This problem was labled in my textbook as a nonlinear equation, but I couldn't find a better spot to put it.
X^2+y^2=3
{
x-y=2
I'm ending up with x= plus or minus sqrt14/2 my textbook has x as plus or minus 1.
Could you explain to me where I'm going wrong?
Thank you!
p.s. I wasn't quite sure how to notate the fact that the two equations went together so I just did the little brackets between them. I'm not a math person, so if you could be very clear and explicit that would be much appreciated. Thanks again!
Found 5 solutions by Fombitz, Theo, KMST, ikleyn, MathTherapy: Answer by Fombitz(32388) (Show Source): Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! 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 https://www.mathsisfun.com/quadratic-equation-solver.html
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.
Answer by KMST(5328) (Show Source): Answer by ikleyn(52775) (Show Source): Answer by MathTherapy(10551) (Show Source):
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