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kati45 [8]
3 years ago
13

Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)

Mathematics
2 answers:
Svet_ta [14]3 years ago
6 0
<h2>Answer:</h2>

\boxed{x=1, \y=-1, \ z=2}

<h2>Step-by-step explanation:</h2>

We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:

Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}

Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}

Step 4: solve for z, then for y, then for x:

\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}

-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}

By substituting y=-1 \ and \ z=2 into the first equation, we get the x. So:

x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}

oksian1 [2.3K]3 years ago
6 0

Answer:

X= 1, y= -1, z= 2

Step-by-step explanation:

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