SOLUTION: 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

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): You can put this solution on YOUR website!
Neither of those answers is correct.
From eq. 2,

Substitute into eq. 1,

Complete the square,

Then use eq. 2 to solve for x.
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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): You can put this solution on YOUR website!
Maybe what you solved and meant to write was

That is the intersection of a larger circle (radius=3)
and the same straight line.

The solution to that system of equations has
.
Answer by ikleyn(52775)   (Show Source): You can put this solution on YOUR website!
.
To see many other solved similar problems/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".

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

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!

I DETEST these types of problems that DON'T have INTEGER solutions.

x - y = 2______x = 2 + y ------- eq (ii)
------- Substituting 2 + y for x in eq (i)

Using the quadratic equation formula, or Completing the Square, solve for y.
You should get 2 different values for y, which are:
Substitute each value for y into any of the 2 ORIGINAL equations (2nd equation is EASIER) to get the 2 corresponding values for x.
There you have it! All done!!
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