Final answer:
To solve the system using the elimination method, we eliminate terms to simplify the algebra. Multiply the second equation by 3 and add it to the first equation to eliminate y. Then, multiply the third equation by -2 and add it to the second equation to eliminate y. Finally, solve for x and y by substituting the values of z back into one of the original equations.
Step-by-step explanation:
To solve the system using the elimination method, we will eliminate terms to simplify the algebra:
1. Multiply the second equation by 3:
-9x - 6y + 15z = -42
2. Add the first equation and the modified second equation:
-6x + 12z = -16
3. Multiply the third equation by -2:
-4x - 8y - 2z = -12
4. Add the third equation and the modified second equation:
-10x + 10z = -28
5. Solve the system of equations by elimination:
-6x + 12z = -16
-10x + 10z = -28
6. Multiply the first equation by -5 and the second equation by 3:
30x - 60z = 80
-30x + 30z = 84
7. Add the equations to eliminate x:
-30z = 164
8. Solve for z:
z = -164/30 = -82/15
9. Substitute the value of z back into one of the original equations to solve for x or y:
-6x + 12(-82/15) = -16
-6x - 984/15 = -16
-6x = -16 + 984/15
-6x = (-240 + 984)/15 = 744/15
x = (744/15) / -6 = -124/15
10. Now substitute the values of x and z back into one of the original equations to solve for y:
3(-124/15) + 2y - 3(-82/15) = -27
-372/15 + 2y + 246/15 = -27
-2y = -27 + 372/15 - 246/15
-2y = (-405 + 372 - 246)/15 = -279/15
y = (-279/15) / -2 = 139/15
Therefore, the solution to the system is x = -124/15, y = 139/15, and z = -82/15.