Solve the simultaneous equations, 2x + 8y - 3z = -1 2x - 3y + 2z = 2 2x + y + z = 5. - Leaving Cert Mathematics - Question 4(a) - 2012

Next, we consider the first and third equations:

1. 2x + 8y - 3z = -1
2. 2x + y + z = 5

Subtracting the third from the first gives:

[(2x + 8y - 3z) - (2x + y + z) = -1 - 5] [7y - 4z = -6].

Now, we have two new equations:

1. 11y - 5z = -3
2. 7y - 4z = -6

Multiply the first equation by 7 and the second by 11:

[77y - 35z = -21] [77y - 44z = -66]

Now, we subtract the second from the first: [9z = 45] [z = 5].

Substituting (z = 5) back into the second equation:

[7y - 4(5) = -6] [7y - 20 = -6] [7y = 14 \Rightarrow y = 2].

Finally, substitute (y = 2) and (z = 5) into the first equation:

[2x + 8(2) - 3(5) = -1] [2x + 16 - 15 = -1] [2x + 1 = -1] [2x = -2 \Rightarrow x = -1].

The solution to the simultaneous equations is: [x = -1, \quad y = 2, \quad z = 5].

We can check our solution by substituting into the original equations:

1. [2(-1) + 8(2) - 3(5) = -1] (True)
2. [2(-1) - 3(2) + 2(5) = 2] (True)
3. [2(-1) + 2 + 5 = 5] (True)

All equations are satisfied.