which ordered pair is in the solution set of the system of equations shown below?
y^2-x^2+32=0
3y-x=0
I did
y^2-3y^2+32=0
y^2-9y^2=-32
8y^2=32
y^2=32/8
y^2=4
y=+/-2
x=3(2)=6
{2,6}
is this right? thank you
1. You got the variables turned around in your answer
2. What about the 2nd value of y ?
Your points are (6,2) and (-6,-2)
see:
www.wolframalpha.com/input/?i=plot+y%5E2-x%5E2%2B32%3D0,+3y-x%3D0
To find the solution set of the given system of equations, you need to solve the equations simultaneously. Let's go through the process step by step.
1. Start with the given system of equations:
y^2-x^2+32 = 0 (Equation 1)
3y-x = 0 (Equation 2)
2. Solve Equation 2 for x:
3y - x = 0
x = 3y (Equation 3)
3. Substitute Equation 3 into Equation 1:
y^2 - (3y)^2 + 32 = 0
y^2 - 9y^2 + 32 = 0
-8y^2 + 32 = 0
-8y^2 = -32
y^2 = 32/8
y^2 = 4
4. Solve for y:
Taking the square root of both sides, we have:
y = ±√4
y = ±2 (Equation 4)
5. Substitute the values of y back into Equation 3 to find x:
For y = 2:
x = 3(2)
x = 6
For y = -2:
x = 3(-2)
x = -6
6. Now you have two ordered pairs:
(x, y) = (6, 2) and (x, y) = (-6, -2)
Therefore, the solution set of the system of equations consists of two ordered pairs: {(6, 2), (-6, -2)}.