help me solve this equation by substitution.
y^2− 4x^2 = 7
9y^2+16x^2 = 323
from the 1st: y^2 = 4x^2 + 7
into the 2nd
9(4x^2 + 7) + 16x^2 = 323
36x^2 + 63 + 16y^2 = 323
52y^2 = 260
y^2 = 5
y = ± √5
sub y^2 = 5 into the 1st:
5 - 4x^2 = 7
-4x^2 = -2
x^2 = 1/2
x = ± 1/√2
4 solutions:
x=√5, y= 1/√2
x = -√5, y = 1/√2
x = √5, y = -1/√2
x = -√5 , y = -1/√2
your graph would show an ellipse intersecting with a hyperbola, with 4 symmetric points of intersection
To solve this system of equations by substitution method, we need to isolate one variable in one equation and substitute it into the other equation. Let's solve for y in the first equation:
y^2 - 4x^2 = 7
Rearranging the equation, we have:
y^2 = 7 + 4x^2
Taking the square root of both sides:
y = ±√(7 + 4x^2)
Now, we substitute this expression for y in the second equation:
9y^2 + 16x^2 = 323
Replacing y with ±√(7 + 4x^2), the equation becomes:
9(±√(7 + 4x^2))^2 + 16x^2 = 323
Simplifying:
9(7 + 4x^2) + 16x^2 = 323
63 + 36x^2 + 16x^2 = 323
Combine like terms:
52x^2 = 260
Divide by 52:
x^2 = 5
Taking the square root of both sides:
x = ±√5
So the possible values for x are x = √5 and x = -√5.
Now, substitute these values of x back into one of the original equations, like the first one:
y^2 - 4x^2 = 7
For x = √5:
y^2 - 4(√5)^2 = 7
y^2 - 4(5) = 7
y^2 - 20 = 7
y^2 = 27
Taking the square root of both sides:
y = ±√27
y = ±3√3
So for x = √5, we have the solutions (x, y) = (√5, 3√3) and (√5, -3√3).
For x = -√5:
y^2 - 4(-√5)^2 = 7
y^2 - 4(5) = 7
y^2 - 20 = 7
y^2 = 27
Taking the square root of both sides:
y = ±√27
y = ±3√3
So for x = -√5, we have the solutions (x, y) = (-√5, 3√3) and (-√5, -3√3).
Therefore, the solution to the system of equations is:
(x, y) = (√5, 3√3), (√5, -3√3), (-√5, 3√3), and (-√5, -3√3).