Solve the system: y^2 - 4x^2 = 4 and y = 2x.
(-1,-2)?
did you plug in your values to see whether they work?
Guess not, since (-1,-2) yields
(-2)^2 - 4(-1) = 4
4 + 4 = 4
Bzzzt
Since y=2x, substitute it into the other equation:
(2x)^2 - 4x^2 = 4
4x^2 - 4x^2 = 4
0 = 4
Bzzzt. No solutions
It feels like too many problems have no solutions! Very frustrating for me. Thank you for checking it!
Yeah, they have tossed a lot of them your way. Expect some on the test, I guess. Always check you proposed answer, if you get one, to make sure it works!
To solve the system of equations, we can substitute the value of y from the second equation into the first equation and solve for x.
Given that y = 2x, we can substitute this into the first equation:
(2x)^2 - 4x^2 = 4
Simplifying:
4x^2 - 4x^2 = 4
0 = 4
Since this equation is not true, there is no value of x that satisfies the system of equations. Therefore, the point (-1,-2) is not a solution to the system.