To find the solution of the system of equations:
-3x - 4y - 3z = -7
2x - 6y + 2z = 3
5x - 2y + 5z = 9
One way to approach this is by using the method of elimination. We can eliminate one variable at a time by manipulating the equations. Let's start by eliminating z.
Multiply the first equation by 2 and the second equation by 3 to create the same coefficient for z:
-6x - 8y - 6z = -14
6x - 18y + 6z = 9
Add the two equations together to eliminate z:
-6x - 8y - 6z + 6x - 18y + 6z = -14 + 9
-26y = -5
y = 5/26
Substitute the value of y back into the first equation to solve for x:
-3x - 4(5/26) - 3z = -7
-3x - 20/26 - 3z = -7
-3x - 10/13 - 3z = -7
-3x - 10/13 - 3z = -7
-3x - 3z = -7 + 10/13
-3x - 3z = -91/13
Next, we need to eliminate z again. Multiply the second equation by 5 and the third equation by 2 to create the same coefficient for z:
10x - 30y + 10z = 15
10x - 4y + 10z = 18
Subtract the two equations to eliminate z:
10x - 30y + 10z - 10x + 4y - 10z = 15 - 18
-26y = -3
y = 3/26
Substitute the value of y back into the second equation to solve for z:
2x - 6(3/26) + 2z = 3
2x - 18/26 + 2z = 3
2x - 9/13 + 2z = 3
2x + 2z = 3 + 9/13
2x + 2z = 48/13
Now we have two equations with two variables:
-3x - 3z = -91/13
2x + 2z = 48/13
Multiply the second equation by -3 to create opposite coefficients for x:
-3(-3x - 3z) = -3(-91/13)
9x + 9z = 273/13
Add this equation to the first equation to eliminate x:
(-3x - 3z) + (9x + 9z) = -91/13 + 273/13
6z = 182/13
z = 182/78
Now substitute the value of z back into the second equation to solve for x:
2x + 2(182/78) = 48/13
2x + 364/78 = 48/13
2x + 182/39 = 48/13
25x + 182 = 192
25x = 10
x = 10/25
x = 2/5
Therefore, the solution to the system of equations is x = 2/5, y = 3/26, and z = 182/78.