Question

solve the system. \\(\

$$\begin{cases} 2x + y + 3z = 20 \\\\ x + 2y - z = -11 \\\\ 3x - 2z = -3 \\end{cases}$$

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

Explanation:

Step 1: Solve the third equation for \(x\)

The third equation is \(3x - 2z=-3\). We can rewrite it as:
\(3x = 2z - 3\)
\(x=\frac{2z - 3}{3}\)

Step 2: Substitute \(x\) into the second equation

The second equation is \(x + 2y - z=-11\). Substitute \(x=\frac{2z - 3}{3}\) into it:
\(\frac{2z - 3}{3}+2y - z=-11\)
Multiply through by 3 to clear the fraction:
\(2z - 3 + 6y - 3z=-33\)
Simplify the left side:
\(-z + 6y - 3=-33\)
\(-z + 6y=-30\)
Let's solve for \(z\):
\(z = 6y + 30\)

Step 3: Substitute \(x\) and \(z\) into the first equation

The first equation is \(2x + y + 3z = 20\). Substitute \(x=\frac{2z - 3}{3}\) and \(z = 6y + 30\) into it. First, substitute \(z = 6y + 30\) into the expression for \(x\):
\(x=\frac{2(6y + 30)-3}{3}=\frac{12y + 60 - 3}{3}=\frac{12y + 57}{3}=4y + 19\)
Now substitute \(x = 4y + 19\) and \(z = 6y + 30\) into the first equation:
\(2(4y + 19)+y + 3(6y + 30)=20\)
Expand the terms:
\(8y + 38 + y + 18y + 90 = 20\)
Combine like terms:
\(27y + 128 = 20\)
Subtract 128 from both sides:
\(27y=20 - 128=-108\)
Divide both sides by 27:
\(y=\frac{-108}{27}=-4\)

Step 4: Find \(z\) using \(y = - 4\)

We had \(z = 6y + 30\). Substitute \(y=-4\):
\(z=6(-4)+30=-24 + 30 = 6\)

Step 5: Find \(x\) using \(z = 6\)

We had \(x=\frac{2z - 3}{3}\). Substitute \(z = 6\):
\(x=\frac{2(6)-3}{3}=\frac{12 - 3}{3}=\frac{9}{3}=3\)

Answer:

\((3, - 4, 6)\)