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bogdanovich [222]
3 years ago
9

2. given 2x2 system X + y - Z=3 2y + 2z = 6 -2y - Z = -2

Mathematics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

Step-by-step explanation:

X + y - Z=3                   ------------------(I)

2y + 2z = 6                ------------------(II)

-2y - Z = -2          ------------------(III)

Multiply equation (III) by 2 and add equation (II) & (III)

(II)         2y + 2z = 6

(III)*2    <u>-4y - 2z = -4</u>        {Now add and z will be eliminated}

            -2y       = 2

                      y = 2/-2

y = -1

Plugin y =-1 in equation (II)

2*(-1) + 2z = 6

-2 +    2z = 6

          2z = 6 + 2

          2z = 8

            z = 8/2

z = 4

Plugin y = -1 & z = 4 in equation(I)

x - 1 - 4  = 3

     x - 5 = 3

          x = 3 + 5

x = 8

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<h2>Answer:</h2>

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

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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}

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