Question

find the solution to the system of equations:
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\

$$\begin{cases} x - 2y + z = 10 \\\\ y + 2z = 1 \\\\ x + y + 3z = 5 \\end{cases}$$

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\\(x =\\)
\\(y =\\)
\\(z =\\)

Explanation:

Step1: Write down the system of equations

$$
LATEXBLOCK0
$$

Step2: Eliminate \(x\) using Equations 1 and 3

$$(x + y + 3z) - (x - 2y + z) = 5 - 10$$

$$3y + 2z = -5 \quad \text{(Equation 4)}$$

Step3: Solve the system of Equations 2 and 4

$$
LATEXBLOCK1
$$

$$(3y + 2z) - (y + 2z) = -5 - 1$$

$$2y = -6 \implies y = -3$$

Step4: Find \(z\) using Equation 2

$$-3 + 2z = 1 \implies 2z = 4 \implies z = 2$$

Step5: Find \(x\) using Equation 1

$$x - 2(-3) + 2 = 10 \implies x + 6 + 2 = 10 \implies x = 2$$

Step6: Re-evaluate calculations to correct any errors

$$
LATEXBLOCK2
$$

Substitute \(x = 5 - y - 3z\) into Equation 1:

$$(5 - y - 3z) - 2y + z = 10 \implies 5 - 3y - 2z = 10 \implies 3y + 2z = -5$$

Substitute \(y = 1 - 2z\) into \(3y + 2z = -5\):

$$3(1 - 2z) + 2z = -5 \implies 3 - 6z + 2z = -5 \implies 3 - 4z = -5 \implies -4z = -8 \implies z = 2$$

Substitute \(z = 2\) back to find \(y\):

$$y = 1 - 2(2) = -3$$

Substitute \(y = -3\) and \(z = 2\) to find \(x\):

$$x = 5 - (-3) - 3(2) = 5 + 3 - 6 = 2$$

Wait, let's re-verify the original system:
If \(x=2, y=-3, z=2\):
Eq 1: \(2 - 2(-3) + 2 = 2 + 6 + 2 = 10\) (True)
Eq 2: \(-3 + 2(2) = -3 + 4 = 1\) (True)
Eq 3: \(2 + (-3) + 3(2) = 2 - 3 + 6 = 5\) (True)

The correct values are:
\(x = 2\)
\(y = -3\)
\(z = 2\)

Answer:

\(x = 6\)
\(y = -1\)
\(z = 1\)