Solving Systems of Equations in Three Variables:
3x+3y-z=-3
-2x-3y+2z=3
x-6y+3z=0
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+3x + 3y - z = -3
-2x - 3y + 2z = 3
x - 6y + 3z = 0
Eq1: 3x + 3y - z = -3
Eq2: -2x - 3y + 2z = 3
Sum: x + 0 + z = 0
z = -x
Eq2: -2x - 3y + 2z = 3
Eq3: x - 6y + 3z = 0
Multiply Eq2 by 6 and Eq3 by (-3):
-12x - 18y + 12z = 18
-3x + 18y - 9z = 0
Sum: -15x + 0 + 3z = 18
Replace z with -x:
-15x-3x = 18
-18x = 18
X = -1
Z = -x = 1.
In Eq1, replace X with -1. and z with 1:
3*(-1) + 3y - 1 = -3
-3 + 3y -1 = -3
3y = 1
Y = 1/3
Solution = (X, Y, Z) = (-1, 1/3, 1).
To solve the system of equations in three variables, we'll use the method called "elimination" or "substitution." Here's how we can proceed:
Step 1: Choose two equations and eliminate one variable.
Let's eliminate the variable "y" from the equations. Looking at the first equation, we can multiply it by 2 to make the coefficient of y equal to -6 (the same as the second equation) or we can multiply the second equation by 3 to make the coefficient of y equal to -6 (the same as the first equation).
We'll choose to multiply the first equation by 2, giving us:
6x + 6y - 2z = -6 (Equation 1')
Now our modified system of equations is:
6x + 6y - 2z = -6 (Equation 1')
-2x - 3y + 2z = 3 (Equation 2)
x - 6y + 3z = 0 (Equation 3)
Step 2: Add Equation 1' to Equation 2 to eliminate "y."
Adding Equation 1' and Equation 2 yields:
6x + 6y - 2z + (-2x - 3y + 2z) = -6 + 3
Simplifying:
6x + (-2x) + 6y + (-3y) - 2z + 2z = -3
Combining like terms:
4x + 3y = -3 (Equation 4)
Now our system is:
4x + 3y = -3 (Equation 4)
-2x - 3y + 2z = 3 (Equation 2)
x - 6y + 3z = 0 (Equation 3)
Step 3: Choose two equations and eliminate the same variable.
Let's eliminate the variable "y" again, but this time we'll use Equation 4 and Equation 2. We can multiply Equation 4 by 2 to make the coefficient of y equal to -6 or multiply Equation 2 by 3 to make the coefficient of y equal to -6.
We'll choose to multiply Equation 4 by 2, giving us:
8x + 6y = -6 (Equation 4')
Now, our modified system is:
8x + 6y = -6 (Equation 4')
-2x - 3y + 2z = 3 (Equation 2)
x - 6y + 3z = 0 (Equation 3)
Step 4: Add Equation 4' to Equation 2 to eliminate "y."
Adding Equation 4' and Equation 2 yields:
8x + 6y + (-2x - 3y + 2z) = -6 + 3
Simplifying:
8x + (-2x) + 6y + (-3y) + 2z = -3
Combining like terms:
6x + 3z = -3 (Equation 5)
Now our system is:
6x + 3z = -3 (Equation 5)
-2x - 3y + 2z = 3 (Equation 2)
x - 6y + 3z = 0 (Equation 3)
Step 5: Choose two equations and eliminate the same variable.
Let's eliminate the variable "x" this time. We can multiply Equation 5 by -2 to make the coefficient of x equal to 6 or multiply Equation 2 by 3 to make the coefficient of x equal to 6.
We'll choose to multiply Equation 5 by -2, giving us:
-12x - 6z = 6 (Equation 5')
Now, our modified system is:
-12x - 6z = 6 (Equation 5')
-2x - 3y + 2z = 3 (Equation 2)
x - 6y + 3z = 0 (Equation 3)
Step 6: Add Equation 5' to Equation 2 to eliminate "x."
Adding Equation 5' and Equation 2 yields:
-12x + (-2x) - 6z + 2z = 6 + 3
Simplifying:
-14x - 4z = 9 (Equation 6)
Now, our system is:
-14x - 4z = 9 (Equation 6)
-2x - 3y + 2z = 3 (Equation 2)
x - 6y + 3z = 0 (Equation 3)
Step 7: We have three linear equations with three variables. We can either apply the substitution method or the elimination method again to solve the system.
For simplicity, let's use the elimination method to solve Equation 6 and Equation 3 to eliminate "z."
Multiply Equation 6 by -3 and Equation 3 by 4 to make the coefficients of z opposite:
42x + 12z = -27 (Equation 6')
4x - 24y + 12z = 0 (Equation 3')
Adding Equation 6' and Equation 3' yields:
42x + 4x + 12z + 12z = -27 + 0
Simplifying:
46x + 24z = -27 (Equation 7)
Now, our system is:
46x + 24z = -27 (Equation 7)
-2x - 3y + 2z = 3 (Equation 2)
x - 6y + 3z = 0 (Equation 3)
Step 8: Solve the resulting system of two equations and two variables.
We now have the following system of equations:
46x + 24z = -27 (Equation 7)
-2x - 3y + 2z = 3 (Equation 2)
x - 6y + 3z = 0 (Equation 3)
To solve this system, we'll choose one variable (either x, y, or z) and solve for it in terms of the remaining variables. Let's choose to solve Equation 3 for x:
x = 6y - 3z (Equation 3')
Substituting Equation 3' into Equation 2 gives:
-2(6y - 3z) - 3y + 2z = 3
Simplifying:
-12y + 6z - 3y + 2z = 3
Combining like terms:
-15y + 8z = 3 (Equation 8)
Now we have the following two equations:
-15y + 8z = 3 (Equation 8)
46x + 24z = -27 (Equation 7)
Step 9: Solve Equation 8 and Equation 7 to find the values of y and z.
To solve this system, we can use the elimination method again. Multiply Equation 8 by 46 and Equation 7 by -15 to make the coefficients of y the same:
-690y + 368z = 138 (Equation 8')
-690x - 360z = 405 (Equation 7')
Adding Equation 8' and Equation 7' yields:
-690y + (-690y) + 368z + (-360z) = 138 + 405
Simplifying:
8z = 543
Dividing both sides by 8:
z = 67.875
Now, we substitute the value of z = 67.875 into Equation 8 to solve for y:
-15y + 8(67.875) = 3
-15y + 543 = 3
-15y = 3 - 543
-15y = -540
y = -540 / -15
y = 36
Finally, we substitute the values of y = 36 and z = 67.875 into Equation 3' to solve for x:
x = 6(36) - 3(67.875)
x = 216 - 203.625
x = 12.375
Therefore, the solution to the system of equations is x = 12.375, y = 36, and z = 67.875.