Question

9. Solve the system algebraically:
x+2y=3z=-11
-2y+z=-1
6z = 30

Answer 1

To solve the system of equations algebraically, rewrite the equations in standard form, eliminate variables, and solve for x, y, and z.

Step-by-step explanation:

To solve the system of equations algebraically:

1. Rewrite the equations in standard form:
   • x + 2y - 3z = -11
   • 0x - 2y + 1z = -1
   • 0x + 0y + 6z = 30
2. Start by eliminating variables. Multiply the second equation by 3 to have -6y + 3z = -3. Add this equation to the first equation:
   • x + 2y - 3z -6y + 3z = -11 - 3
   • x - 4y = -14
3. Next, multiply the second equation by 2 to have 0x - 4y + 2z = -2. Add this equation to the third equation:
   • 0x + 0y + 6z - 4y + 2z = 30 - 2
   • -2y + 8z = 28
4. The system of equations is now reduced to:
   • x - 4y = -14
   • -2y + 8z = 28
   • 0x + 0y + 6z = 28
5. Now, focus on the equations involving y and z. Multiply the third equation by 4/6 to have 0x + 0y + 4z = 20/6. Add this equation to the second equation:
   • -2y + 8z + 0y + 4z = 28 + 20/6
   • 12z = 76/6
   • 2z = 76/36
   • z = 19/36
6. Substitute the value of z into the second equation:
   • -2y + 8(19/36) = 28
   • -2y = 28 - 152/36
   • -2y = -20/36
   • y = 10/36
   • y = 5/18
7. Substitute the values of y and z into the first equation:
   • x - 4(5/18) = -14
   • x = -14 + 20/18
   • x = -14 + 10/9
   • x = -14/1 + 10/9
   • x = -126/9 + 10/9
   • x = -116/9
8. The solution to the system of equations is:
   • x = -116/9
   • y = 5/18
   • z = 19/36

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