Find the exact solution(s) of each system of equations?
x^2+y^2=25 and 9y=4x^2
a.(4,3),(-4,3)
b.(3,4),(3,-4)
c.(4,3),(4,-3)
d.(3,4),(-3,4)
I had d but im not sure if that is right.
To find the exact solution(s) of the system of equations, let's start by rearranging the second equation to solve for y:
9y = 4x^2
Divide both sides by 9:
y = (4/9)x^2
Now we can substitute this value of y into the first equation:
x^2 + [(4/9)x^2] = 25
Combine like terms:
(1 + 4/9)x^2 = 25
(9/9 + 4/9)x^2 = 25
(13/9)x^2 = 25
Multiply both sides by 9/13:
x^2 = (25 * 9) / 13
x^2 = 225 / 13
Take the square root of both sides:
x = ± √(225 / 13)
Simplifying further:
x = ± (15 / √13)
So the x-values for the possible solutions are ± (15 / √13).
Now let's substitute these values of x back into the second equation to find the corresponding y-values.
When x = (15 / √13):
y = (4/9)(15 / √13)^2
Simplify:
y = (4/9)(225 / 13)
y = 900 / (9 * 13)
y = 100 / 13
So one solution is (x, y) = (15 / √13, 100 / 13).
When x = -(15 / √13):
y = (4/9)(-(15 / √13))^2
Simplify:
y = (4/9)(225 / 13)
y = 900 / (9 * 13)
y = 100 / 13
So the other solution is (x, y) = (-(15 / √13), 100 / 13).
Therefore, the correct answer is c.(4,3),(4,-3), which matches the solutions we obtained.
Let's see if d satisfies the equations.
x=3 and y = 4
equation 1.
3^2 + 4^2 = 25
9 + 15 = 25 OK on this one.
x = -3 and y = 4
equation 2.
9*(-3) = 4*(4^2)
-27 = 4*16
-27=64 so d isn't right. The first set satisfied the first equation but the second set did not satify the second equation.