Find the exact solutions of the systems of equation x^2+2y^2=18 and x=2y
(2y)^2 + 2 y^2 = 18
4 y^2 + 2 y^2 = 18
6 y^2 = 18
y^2 = 3
y = +sqrt 3 or -sqrt 3
x^2 + 2 *3 = 18
x^2 = 12
x = + or - 2 sqrt 3
20=+or-2 sqrt 3=5
To find the exact solutions for the system of equations, we can substitute the value of x from the second equation into the first equation.
Given:
x = 2y
Substituting this value of x into the first equation:
(2y)^2 + 2y^2 = 18
Simplifying the equation:
4y^2 + 2y^2 = 18
6y^2 = 18
y^2 = 18/6
y^2 = 3
Taking the square root of both sides:
y = ±√3
Now we substitute the values of y back into the second equation to find the corresponding values of x:
For y = √3:
x = 2(√3) = 2√3
For y = -√3:
x = 2(-√3) = -2√3
Therefore, the exact solutions to the system of equations x^2 + 2y^2 = 18 and x = 2y are:
(x, y) = (2√3, √3) and (-2√3, -√3)