Given 2x^2+3y^2=35.Find the number of ordered pairs (x,y) where (x,y) are integers satisfying the given condition.
To find the number of ordered pairs (x,y) where x and y are integers such that 2x^2 + 3y^2 = 35, we can start by noticing that the equation is that of an ellipse.
First, we need to rewrite the equation in the form of (x/a)^2 + (y/b)^2 = 1:
Dividing by 35 on both sides:
(2x^2)/35 + (3y^2)/35 = 1
(2x^2)/35 = 1 - (3y^2)/35
(2x^2)/35 = (35 - 3y^2)/35
2x^2 = 35 - 3y^2
2x^2 + 3y^2 = 35
Now we can see that a^2 = 35/2 = 17.5 and b^2 = 35/3 = 11.66 which are not integers. This means that there are no integer solutions for this equation. So the number of ordered pairs (x,y) where x and y are integers satisfying the given condition is 0.