If 3xy = 9xz = 12yz = 1, what is the value of x, y, and z?
This is just a set of 3 equations:
3xy=1
9xz=1
12yz=1
⇒(18xyz)^2=1
⇒ xyz = ± 1/18
⇒ (x,y,z) = ±(2/3,1/2,1/6)
PLEASE TELL ME HOW DID THE SOLUTION CAME UP WITH xyz = ± 1/18 and (x,y,z) = ±(2/3,1/2,1/6. THANKS!
3 x y = 9 x z = 12 y z = 1 Divide by 3
x y = 3 x z = 4 y z = 1 / 3
x y = 1 / 3 Divide both sides by x
y = 1 / ( 3 x )
3 x z = 1 / 3 Divide both sides by 3 x
z = 1 / ( 9 x )
4 y z = 1 / 3
4 * 1 / ( 3 x ) * 1 / ( 9 x ) = 1 / 3
4 * 1 / ( 27 x ^ 2 ) = 1 / 3
4 / 27 x ^ 2 = 1 / 3 Multiply both sides by 27 x ^ 2
4 = ( 1 / 3 ) * 27 x ^ 2
4 = 27 x ^ 2 / 3
4 = 9 x ^ 2
9 x ^ 2 = 4 Divide both sides by 9
x ^ 2 = 4 / 9
x = + OR - sqrt ( 4 / 9 )
x = + OR - 2 / 3
For x = - 2 / 3
y = 1 / ( 3 x )
y = 1 / [ 3 * ( - 2 ) / 3 ]
y = 1 / - 2
y = - 1 / 2
z = 1 / ( 9 x )
z = 1 / [ 9 * ( - 2 ) / 3 ]
z = 1 / [ ( 9 / 3 ) * - 2 )
z = 1 / [ 3 * ( - 2 ) ]
z = 1 / - 6
z = - 1 / 6
For x = 2 / 3
y = 1 / ( 3 x )
y = 1 / ( 3 * 2 / 3 )
y = 1 / 2
z = 1 / ( 9 x )
z = 1 / ( 9 * 2 / 3 )
z = 1 / [ ( 9 / 3 ) * 2 )
z = 1 / ( 3 * 2 )
z = 1 / 6
You have two set of solutions :
x = - 2 / 3 , y = - 1 / 2 , z = - 1 / 6
and
x = 2 / 3 , y = 1 / 2 , z = 1 / 6
Well, solving this system of equations can be quite tricky, but let me try to explain it in a funny way!
Imagine you're at a circus and you're about to perform a magic trick. You have three boxes, and you need to figure out what's inside each box. The only clue you have is that the sum of the numbers inside each box is 1.
You start by labeling the boxes as x, y, and z. So, the first equation, 3xy = 1, tells you that the sum of what's inside boxes x and y is 1/3.
Next, the second equation, 9xz = 1, tells you that the sum of what's inside boxes x and z is 1/9.
Lastly, the third equation, 12yz = 1, tells you that the sum of what's inside boxes y and z is 1/12.
Now, what you want to find is the values of x, y, and z. To do that, you have to combine the information from all three equations.
Think of it this way: if you add up the contents of all three boxes together, it should still sum up to 1. So you can write an equation like this: (x + y + z)² = 1.
Expanding the equation gives you x² + y² + z² + 2xy + 2xz + 2yz = 1.
Now, substitute the values you found for the sums of the boxes from the original equations.
You have (1/3)² + (1/9)² + (1/12)² + 2(1/3)xy + 2(1/9)xz + 2(1/12)yz = 1.
Simplifying that equation, you get 1/9 + 1/81 + 1/144 + 2xy/3 + 2xz/9 + 2yz/12 = 1.
Since the sum of the squared numbers is 1/9 + 1/81 + 1/144, and since we know that 2xy/3 = 1/3, 2xz/9 = 1/9, and 2yz/12 = 1/12, you can combine those terms.
That gives you 1/9 + 1/81 + 1/144 + 1/3 + 1/9 + 1/12 = 1.
Now, combine the fractions by finding a common denominator. That will give you (144 + 16 + 9 + 48 + 16 + 12)/144 = 1.
Simplifying that further, you have 245/144 = 1.
Multiplying both sides of the equation by 144 gives you 245 = 144.
Wait a minute, that's not right! In fact, that's mathematically impossible!
So, it turns out that there is no real solution to this system of equations. However, if you consider complex numbers, you can find an imaginary solution.
In short, xy = 1/3, xz = 1/9, yz = 1/12. By combining these equations, you can solve for xyz and find that xyz = ±1/18.
As for the values of x, y, and z, you can substitute xyz = ±1/18 into the original equations and solve for each variable.
That's how we got the values (x, y, z) = ±(2/3, 1/2, 1/6).
I hope this explanation brought a smile to your face! If you have any more questions, feel free to ask!
To find the value of x, y, and z, we need to solve the system of equations:
3xy = 9xz = 12yz = 1
First, let's simplify the equations:
3xy = 1
9xz = 1
12yz = 1
Looking at these equations, we can see that the right side of each equation is the same, which is 1.
Now, let's solve for x in terms of y and z:
From the first equation, 3xy = 1, we can rearrange it as x = 1/(3y).
Now, let's substitute this expression for x in the other two equations:
9(1/(3y))z = 1
12yz = 1
Simplifying the equation 9(1/(3y))z = 1:
(3z)/(y) = 1/9
Cross-multiplying, we get:
3z = (y)(1/9)
3z = y/9
z = y/27
Now, let's substitute this expression for z in the third equation:
12y(y/27) = 1
12y^2/27 = 1
12y^2 = 27
y^2 = 27/12
y^2 = 9/4
Taking the square root of both sides, we get:
y = ± √(9/4)
y = ± (3/2)
Substituting this value of y back into the expression for z:
z = y/27
z = (± (3/2))/27
z = ± (1/18)
Lastly, we substitute the values of y and z back into the expression for x:
x = 1/(3y)
x = 1/(3(± (3/2)))
x = ± (2/3)
Therefore, the solutions for x, y, and z are:
(x, y, z) = ± (2/3, 3/2, 1/18)
or
(x, y, z) = ± (2/3, -3/2, -1/18)
Please note that we have ± sign because both positive and negative values satisfy the equations.
To solve this system of equations, we can use the method of substitution.
First, let's isolate one variable in terms of the other two variables. We can start by isolating x in terms of y and z from the first equation:
3xy = 1
Divide both sides by 3y:
x = 1/(3y)
Now we can substitute this value of x into the second equation:
9xz = 1
Replace x with 1/(3y):
9(1/(3y))z = 1
3z/y = 1
Multiply both sides by y/3:
z = y/3
Now we have expressions for x and z in terms of y. We can substitute these into the third equation:
12yz = 1
Replace z with y/3 and x with 1/(3y):
12y(y/3) = 1
4y^2 = 1
y^2 = 1/4
y = ±√(1/4)
y = ±(1/2)
Now that we have found the possible values for y, we can substitute these back into the equations to find the corresponding values for x and z.
If y = 1/2, substituting into the equation x = 1/(3y):
x = 1/(3(1/2))
x = 1/(3/2)
x = 2/3
And substituting into the equation z = y/3:
z = (1/2)/3
z = 1/6
Therefore, one solution is (x, y, z) = (2/3, 1/2, 1/6).
If y = -1/2, the process is the same, but with negative values:
x = 2/(-3)
x = -2/3
z = (-1/2)/3
z = -1/6
Therefore, the second solution is (x, y, z) = (-2/3, -1/2, -1/6).
Hence, the values of x, y, and z are ±(2/3, 1/2, 1/6).