Solve the following systems of equations
4x^2 - 2y^2 = 4
x^2 + y^2 = 10
Rearrange both equations so that y^2 is the subjects of the equation.
Then put them together to solve for x
Give that a go and post your working. Then, if you need it, we can help you more. :)
Another option: re arrange the second equation to this: y^2 = 10 - x^2 Now substitute THIS value of y^2 into the FIRST equation and solve for 'x'.....once you have an 'x' value(s)....sub this value of x into one of the original equations to calculate the corresponding 'y' value(s).....
let us know what you find!
A third option would be to graph the two equations and look for the points that they intersect...... try that too !
4x^2 - 2y^2 = 4 ⇒ 2x^2 -y^2 = 2 (1)
x^2 + y^2 = 10 (2)
Add (1) and (2) to get
3x^2 = 12 divide both sides by 3
x^2 = 4 take both roots
x = ±√4
x = 2 and x = -2
And when x is either of these values, we can find y as
(2)^2 + y^2 = 10
4 + y^2 = 10
y^2 = 6
y = ±√6
So.....we have these 4 solutions
(2, √6) (2, -√6) ( -2, √6) (-2, - √6 )