Which ordered pair is not in the inverse of the relation given by xy^2-3x = 12?
(A) (0, -4)
(B) (4, 1)
(C) (3, 2)
(D) (2, 12)
(E) (1, -6)
y x^2 - 3 y = 12
y (x^2 - 3) = 12
y = 12 / (x^2-3)
if x = 0 , y = -4 ok
if x = 4 , y = 12/13 oops
etc
Well, let's see here. To find the inverse of the relation, we need to switch the x and y variables and solve for y. So, let's solve the equation xy^2 - 3x = 12 for y.
xy^2 - 3x = 12
Now, we can switch the x and y variables:
yx^2 - 3y = 12
Now, let's rearrange the equation to solve for y:
yx^2 - 3y - 12 = 0
This is a quadratic equation in terms of y. We can solve it using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
Plug in the values for a, b, and c:
a = x
b = -3
c = -12
Now, let's find the value of y for each ordered pair given:
(A) (0, -4)
Plugging in x = 0:
y = (0 ± √((-3)^2 - 4(0)(-12))) / (2(0))
This becomes y = ± √(9) / 0, which is undefined. So, (0, -4) is not in the inverse of the relation.
(B) (4, 1)
Plugging in x = 4:
y = (4 ± √((-3)^2 - 4(4)(-12))) / (2(4))
This becomes y = (4 ± √(105)) / 8
So, (4, 1) is in the inverse of the relation.
(C) (3, 2)
Plugging in x = 3:
y = (3 ± √((-3)^2 - 4(3)(-12))) / (2(3))
This becomes y = (3 ± √(129)) / 6
So, (3, 2) is in the inverse of the relation.
(D) (2, 12)
Plugging in x = 2:
y = (2 ± √((-3)^2 - 4(2)(-12))) / (2(2))
This becomes y = (2 ± √(171)) / 4
So, (2, 12) is in the inverse of the relation.
(E) (1, -6)
Plugging in x = 1:
y = (1 ± √((-3)^2 - 4(1)(-12))) / (2(1))
This becomes y = (1 ± √(69)) / 2
So, (1, -6) is in the inverse of the relation.
Therefore, the ordered pair not in the inverse of the relation is (A) (0, -4).
To find the inverse of the given relation, we need to solve the equation xy^2 - 3x = 12 for x and y.
Step 1: Start with the given equation: xy^2 - 3x = 12
Step 2: Subtract 12 from both sides: xy^2 - 3x - 12 = 0
Step 3: Factor out x: x(y^2 - 3) - 12 = 0
Step 4: Solve for x: x = 12 / (y^2 - 3)
Step 5: Swap x and y to find the inverse relation: y = 12 / (x^2 - 3)
Now, let's check if each of the given ordered pairs are in the inverse relation.
(A) (0, -4): substituting x = 0 and y = -4 into the inverse relation, we get -4 = 12 / (-3), which is not true.
(B) (4, 1): substituting x = 4 and y = 1 into the inverse relation, we get 1 = 12 / (4^2 - 3), which is true.
(C) (3, 2): substituting x = 3 and y = 2 into the inverse relation, we get 2 = 12 / (3^2 - 3), which is true.
(D) (2, 12): substituting x = 2 and y = 12 into the inverse relation, we get 12 = 12 / (2^2 - 3), which is not true.
(E) (1, -6): substituting x = 1 and y = -6 into the inverse relation, we get -6 = 12 / (1^2 - 3), which is true.
Therefore, the ordered pair that is not in the inverse of the relation is (A) (0, -4).
To find the inverse of the relation given by xy^2 - 3x = 12, we need to solve this equation for y in terms of x and swap the x and y variables.
First, let's solve the equation xy^2 - 3x = 12 for y:
xy^2 = 12 + 3x
y^2 = (12 + 3x) / x
y = ± √((12 + 3x) / x)
Now, we swap the x and y variables to obtain the inverse relation:
x = ± √((12 + 3y) / y)
Now, let's check each ordered pair given in the options to see which one is not in the inverse relation:
(A) (0, -4):
x = 0, y = -4
Checking the inverse relation: 0 = ± √((12 + 3(-4)) / (-4))
0 = ± √(12 / -4)
Since the square root of a negative number is not real, this ordered pair (0, -4) is not in the inverse relation.
(B) (4, 1):
x = 4, y = 1
Checking the inverse relation: 4 = ± √((12 + 3(1)) / 1)
4 = ± √(15 / 1)
Both the positive and negative square roots of 15 are real numbers, so this ordered pair (4, 1) is in the inverse relation.
(C) (3, 2):
x = 3, y = 2
Checking the inverse relation: 3 = ± √((12 + 3(2)) / 2)
3 = ± √(18 / 2) = ± √9
The square root of 9 is ±3, so this ordered pair (3, 2) is in the inverse relation.
(D) (2, 12):
x = 2, y = 12
Checking the inverse relation: 2 = ± √((12 + 3(12)) / 12)
2 = ± √(12 / 12) = ± √1
The square root of 1 is ±1, so this ordered pair (2, 12) is in the inverse relation.
(E) (1, -6):
x = 1, y = -6
Checking the inverse relation: 1 = ± √((12 + 3(-6)) / (-6))
1 = ± √(0 / -6) = ± √0
The square root of 0 is 0, so this ordered pair (1, -6) is in the inverse relation.
Based on the results, the ordered pair that is not in the inverse of the relation is (A) (0, -4).