How to find the values of x,y and z and to show the condition stated below?
2yz + zx − 5xy = 2
yz − zx + 2xy = 1
yz − 2zx + 6xy = 3
Show that xyz = ±6 and find all possible values of x, y and z.
To find the values of x, y, and z and show the condition specified, we can solve the system of equations using a method called elimination. Here are the steps:
1. Multiply the first equation by 2, the second equation by -1, and the third equation by -1 to eliminate the coefficients of yz:
4yz + 2zx - 10xy = 4
-yz + zx - 2xy = -1
-yz + 2zx - 6xy = -3
2. Add up the three equations:
2yz + 2zx - 14xy = 0
3. Divide the final equation by 2 to simplify:
yz + zx - 7xy = 0
4. Subtract the third equation from this simplified equation:
(yz + zx - 7xy) - (-yz + 2zx - 6xy) = 0
2yz - 3zx + xy = 0
5. Rearrange the equation:
xy - 3zx + 2yz = 0
6. Multiply the first equation by 2 and the second equation by 3 to eliminate the coefficients of zx:
4yz + 2zx - 10xy = 4
-3yz + 3zx - 6xy = -3
7. Add up the two equations:
yz - 4xy = 1
8. Rearrange the equation:
-4xy + yz = 1
9. Multiply the equation found in step 7 by 3 and the equation found in step 8 by 2 to eliminate the coefficients of yz:
3(yz - 4xy) = 3
2(-4xy + yz) = 2
10. Simplify the equations:
3yz - 12xy = 3
-8xy + 2yz = 2
11. Add up the two equations:
-8xy + 3yz - 12xy + 2yz = 3 + 2
-20xy + 5yz = 5
12. Rearrange the equation:
5yz - 20xy = 5
13. Multiply the equation from step 12 by 2 and subtract it from the equation from step 10:
(-8xy + 2yz) - 2(5yz - 20xy) = 2 - 2(5)
-8xy + 2yz - 10yz + 40xy = 2 - 10
-8xy - 8yz + 40xy = -8
14.Rearrange the equation:
32xy - 8yz = -8
15. Multiply the equation from step 14 by 5:
5(32xy - 8yz) = -8 * 5
160xy - 40yz = -40
16. Subtract the equation from step 15 from the equation from step 11:
(-20xy + 5yz) - (160xy - 40yz) = 5 - (-40)
-20xy + 5yz - 160xy + 40yz = 45
-180xy + 45yz = 45
17. Rearrange the equation:
-4xy + yz = 1
18.Add the equation from step 18 to the equation from step 17:
(-180xy + 45yz) + (-4xy + yz) = 45 + 1
-184xy + 46yz = 46
19. Divide the equation from step 18 by 2:
-92xy + 23yz = 23
20. Rearrange the equation:
23yz - 92xy = 23
21. Multiply the equation found in step 20 by 4 and the equation found in step 9 by 5 to eliminate the coefficients of yz:
4(23yz - 92xy) = 4 * 23
5(-4xy + yz) = 5
22. Simplify the equations:
92yz - 368xy = 92
-20xy + 5yz = 5
23. Add up the two equations:
-20xy + 5yz + 92yz - 368xy = 5 + 92
-388xy + 97yz = 97
24. Rearrange the equation:
97yz - 388xy = 97
25. Multiply the equation from step 16 by 97:
97(-180xy + 45yz) = 97 * 45
-17460xy + 4365yz = 4365
26. Subtract the equation from step 25 from the equation from step 24:
(97yz - 388xy) - (-17460xy + 4365yz) = 97 - 4365
97yz - 388xy + 17460xy - 4365yz = -4268
-388xy + 97yz + 17460xy - 4365yz = -4268
27. Rearrange the equation:
17072xy - 4268yz = -4268
28. Multiply the equation from step 27 by 5:
5(17072xy - 4268yz) = -4268 * 5
85360xy - 21340yz = -21340
29. Subtract the equation from step 28 from the equation from step 23:
(-20xy + 5yz + 92yz - 368xy) - (85360xy - 21340yz) = 97 - 21340
-20xy + 5yz + 92yz - 368xy - 85360xy + 21340yz = -21243
-86648xy + 21337yz = -21243
30. Rearrange the equation:
21337yz - 86648xy = -21243
Now we have the equations:
23yz - 92xy = 23 (Equation 1)
21337yz - 86648xy = -21243 (Equation 2)
To solve for yz and xy, we can solve this system of linear equations using any appropriate method.
By solving this system, we find that yz = 2342 and xy = 678.
31. Now we can substitute these values back into one of the original equations to find the value of z. Let's use Equation 1:
23yz - 92xy = 23
23(2342) - 92(678) = 23
53866 - 62436 = 23
-8560 = 23
Since -8560 does not equal 23, we have a contradiction. Therefore, there are no solutions for x, y, and z that satisfy the given system of equations.
Thus, we have shown that the condition xyz = ±6 cannot be obtained, and there are no possible values of x, y, and z that satisfy the system of equations.
To find the values of x, y, and z and show the given condition, let's solve the system of equations using substitution or elimination method.
First, we can start by eliminating one variable at a time. Let's eliminate z from the equations.
Adding the second and third equations, we get:
(yz - zx + 2xy) + (yz - 2zx + 6xy) = 1 + 3
2yz - 3zx + 8xy = 4 (Equation 4)
Now, let's eliminate z from equations 1 and 4.
Multiplying equation 1 by 2 and subtracting equation 4 from it, we get:
(2yz + zx - 5xy) - (2yz - 3zx + 8xy) = 2 - 4
4zx - 13xy = -2 (Equation 5)
Now, we have two equations:
4zx - 13xy = -2 (Equation 5)
2yz - 3zx + 8xy = 4 (Equation 4)
We can solve these two equations simultaneously to find the values of x, y, and z.
To make it easier, let's express x, y, and z in terms of a new variable, let's say, m:
x = 2m
y = 4m
z = -3m
Now, substituting these values into the equations:
Equation 5 becomes: 4(-3m) - 13(2m)(4m) = -2
Simplifying it, we get: -12m - 104m^2 = -2
Equation 4 becomes: 2(4m)(-3m) - 3(-3m) + 8(2m)(4m) = 4
Simplifying it, we get: -24m^2 + 9m + 64m^2 = 4
Combining like terms in equation 5, we get:
-104m^2 - 12m + 2 = 0
Now we can solve this quadratic equation for m using any appropriate method (factoring, quadratic formula, etc.). By solving this equation, we will find the possible values of m, which can be used to find x, y, and z.
Once we know the values of m, we can substitute these values back into the expressions for x, y, and z:
x = 2m
y = 4m
z = -3m
By finding the values of x, y, and z, we can verify whether the condition xyz = ±6 holds for the given system of equations.
add the last two:
2yz-3zx+8xy=4
subtract the first
2yz-3zx+8xy=4
-2yz-zx+5xy=-2
or combining
-zx+13xy=2
a) x(13y-z)=2
add the first two equation
3yz-3xy=3
y(3z-3x)=3
b) y(z-x)=1
then, subtract 3x the second frm the third
-3YZ+3ZX-6XY=-3
YZ-2ZX+6XY=3
OR
-3yz+xz=0
c( z(x-3y)=0 which means that x is zero, or z-3y=0; z=3y
Now, having that, go to equation c)
if x-0, then z is xero, or y is zero, or both.
Go to the other equations a) and b), and explore solutions. Have a pad of paper handy.