Solve the system, 4x^2 + y^2 = 20 and y = 4x.
(square root of 10/2, 2 square root of 10)?
Hmmm. I get (-1,-4) and (1,4)
y=4x, so
4x^2 + (4x)^2 = 20
4x^2 + 16x^2 = 20
20x^2 = 20
x = ±1
so y=±4
Thank you
To solve the system, we have two equations:
1) 4x^2 + y^2 = 20
2) y = 4x
We can substitute the value of y from the second equation (y = 4x) into the first equation (4x^2 + y^2 = 20) to get rid of the variable y.
Substituting y = 4x into the first equation gives us:
4x^2 + (4x)^2 = 20
4x^2 + 16x^2 = 20
20x^2 = 20
x^2 = 1
Now, we take the square root of both sides of the equation to solve for x:
√(x^2) = ±√1
x = ±1
Since we now have the value of x, we can substitute it back into the second equation (y = 4x) to find the corresponding values of y.
For x = 1: y = 4(1) = 4
For x = -1: y = 4(-1) = -4
Therefore, the solutions are (1, 4) and (-1, -4).