Question

solve the system.
\

$$\begin{cases}x + 2y - 4z = 4 \\\\ 2x - y + 2z = 3 \\\\ 3x - 8z = -2\\end{cases}$$

enter your answer as an ordered triple (x, y, z).
(?, \square, \square)

Explanation:

Step1: Solve the third equation for \(x\)

From \(3x - 8z=-2\), we can express \(x\) in terms of \(z\).
Add \(8z\) to both sides: \(3x = 8z - 2\)
Divide both sides by \(3\): \(x=\frac{8z - 2}{3}\)

Step2: Substitute \(x\) into the second equation

The second equation is \(2x - y + 2z = 3\). Substitute \(x=\frac{8z - 2}{3}\) into it:
\(2\times\frac{8z - 2}{3}-y + 2z = 3\)
Multiply through by \(3\) to eliminate the denominator: \(2(8z - 2)-3y + 6z = 9\)
Expand: \(16z - 4 - 3y + 6z = 9\)
Combine like terms: \(22z - 4 - 3y = 9\)
Rearrange to solve for \(y\): \(-3y=9 + 4 - 22z\)
\(-3y = 13 - 22z\)
Divide by \(-3\): \(y=\frac{22z - 13}{3}\)

Step3: Substitute \(x\) into the first equation

The first equation is \(x + 2y - 4z = 4\). Substitute \(x=\frac{8z - 2}{3}\) and \(y=\frac{22z - 13}{3}\) into it:
\(\frac{8z - 2}{3}+2\times\frac{22z - 13}{3}-4z = 4\)
Multiply through by \(3\): \(8z - 2 + 2(22z - 13)-12z = 12\)
Expand: \(8z - 2 + 44z - 26 - 12z = 12\)
Combine like terms: \((8z + 44z - 12z)+(-2 - 26)=12\)
\(40z - 28 = 12\)
Add \(28\) to both sides: \(40z = 12 + 28\)
\(40z = 40\)
Divide by \(40\): \(z = 1\)

Step4: Find \(x\) when \(z = 1\)

Substitute \(z = 1\) into \(x=\frac{8z - 2}{3}\):
\(x=\frac{8\times1 - 2}{3}=\frac{6}{3}=2\)

Step5: Find \(y\) when \(z = 1\)

Substitute \(z = 1\) into \(y=\frac{22z - 13}{3}\):
\(y=\frac{22\times1 - 13}{3}=\frac{9}{3}=3\)

Answer:

\((2, 3, 1)\)