What is the solution set for the following system of equations: x + 2y + 3z = 16, x + 2y + 2z = 10, 5x + 3y = 4?

What is the solution set for the following system of equations: x + 2y + 3z = 16, x + 2y + 2z = 10, 5x + 3y = 4?

• A. (2, -2, 6)
• B. (1, 0, 5)
• C. (3, 2, 4)
• D. (2, 2, 2)

Answer: (A)

To find the solution set for the system of equations, we need to solve them simultaneously. The first two equations are: 1. \(x + 2y + 3z = 16\) 2. \(x + 2y + 2z = 10\) Subtracting the second equation from the first gives us an equation involving \(z\): \[ (x + 2y + 3z) - (x + 2y + 2z) = 16 - 10 \] \[ z = 6 \] Now that we have \(z = 6\), we can substitute this value back into either of the original equations to solve for \(x\) and \(y\). Using the second equation: \[ x + 2y + 2(6) = 10 \] \[ x + 2y + 12 = 10 \] \[ x + 2y = -2 \tag{Equation 3} \] Next, we take the third equation from the system: \[ 5x + 3y = 4 \tag{Equation 4} \] Now we solve the system formed by

To find the solution set for the system of equations, we need to solve them simultaneously.

The first two equations are:

1. (x + 2y + 3z = 16)

2. (x + 2y + 2z = 10)

Subtracting the second equation from the first gives us an equation involving (z):

[

(x + 2y + 3z) - (x + 2y + 2z) = 16 - 10

]

[

z = 6

]

Now that we have (z = 6), we can substitute this value back into either of the original equations to solve for (x) and (y). Using the second equation:

[

x + 2y + 2(6) = 10

]

[

x + 2y + 12 = 10

]

[

x + 2y = -2 \tag{Equation 3}

]

Next, we take the third equation from the system:

[

5x + 3y = 4 \tag{Equation 4}

]

Now we solve the system formed by