Solve the system of equations: 8x + 4y + 27 = 3, 2x - 3y - 2 = -3, x + 2y + z = -4.

Steps to Solve

1. Write the system of equations clearly

From the given information, we can extract the following equations:

[ \begin{align*} (1) & \quad 8x + 4y + 27 = 3 \ (2) & \quad 2x - 3y - 2 = -3 \ (3) & \quad x + 2y + z = 2 \end{align*} ]

2. Rearrange each equation

We will rearrange the first two equations to isolate the constants on the right side:

[ \begin{align*} (1) & \quad 8x + 4y = 3 - 27 \Rightarrow 8x + 4y = -24 \ (2) & \quad 2x - 3y = -3 + 2 \Rightarrow 2x - 3y = -1 \ (3) & \quad x + 2y + z = 2 \text{ (remains the same)} \end{align*} ]

3. Simplify the equations

The equations now look like:

[ \begin{align*} (1) & \quad 8x + 4y = -24 \ (2) & \quad 2x - 3y = -1 \ (3) & \quad x + 2y + z = 2 \end{align*} ]

4. Solve the equations using substitution or elimination

Start with equations (1) and (2) to eliminate one variable. We can solve (2) for $x$:

From (2):

[ x = \frac{-1 + 3y}{2} ]

5. Substitute x in (1)

Now substitute $x$ in equation (1):

[ 8\left(\frac{-1 + 3y}{2}\right) + 4y = -24 ]

Multiply through by 2 to clear the fraction:

[ 8(-1 + 3y) + 8y = -48 ]

6. Solve for y

This simplifies to:

[ -8 + 24y + 8y = -48 ]

Combining terms gives:

[ 32y = -40 ]

Thus,

[ y = -\frac{40}{32} = -\frac{5}{4} ]

7. Substitute y back to find x

Now use $y = -\frac{5}{4}$ in the equation for $x$ from step 4:

[ x = \frac{-1 + 3(-\frac{5}{4})}{2} = \frac{-1 - \frac{15}{4}}{2} = \frac{-\frac{4}{4} - \frac{15}{4}}{2} = \frac{-\frac{19}{4}}{2} = -\frac{19}{8} ]

8. Substituting x and y to find z

Now substitute $x$ and $y$ into equation (3):

[ -\frac{19}{8} + 2(-\frac{5}{4}) + z = 2 ]

Calculate $2(-\frac{5}{4}) = -\frac{10}{4} = -\frac{5}{2} = -\frac{20}{8}$.

So:

[ -\frac{19}{8} - \frac{20}{8} + z = 2 ] [ z = 2 + \frac{39}{8} = \frac{16}{8} + \frac{39}{8} = \frac{55}{8} ]

Thus, the solutions are:

[ \begin{align*} x & = -\frac{19}{8} \ y & = -\frac{5}{4} \ z & = \frac{55}{8} \end{align*} ]