Question

Solve the system of equations.

Answer 1

$x = \frac{ 419 }{ 113 } ~,~y = -\frac{ 208 }{ 113 } ~,~z = \frac{ 21 }{ 113 }$

Explanation:

We have the following system of three linear equations:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\5~ x&+~~8~ y&+~~~~ z&~=~4\\4~ x&+~~5~ y&+~~2~ z&~=~6\end{array}$

Let's use elimination method in order to get the solution of this system of equation, so let's solve this step by step.

Step 1: Multiply first equation by $-5/2$ and add the result to the second equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\4~ x&+~~5~ y&+~~2~ z&~=~6\end{array}$

Step 2: Multiply first equation by −2 and add the result to the third equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\&-~~~3~ y&-~~~62~ z&~=~-6\end{array}$

Step 3: Multiply second equation by −32 and add the result to the third equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\&&+~~\frac{ 113 }{ 2 }~ z&~=~\frac{ 21 }{ 2 }\end{array}$

Step 4: solve for z.

$\begin{aligned} \frac{ 113 }{ 2 } ~ z & = \frac{ 21 }{ 2 } \\ z & = \frac{ 21 }{ 113 } \end{aligned}$

Step 5: solve for y.

$\begin{aligned}-2y-79z &= -11\\-2y-79\cdot \frac{ 21 }{ 113 } &= -11\\y &= -\frac{ 208 }{ 113 } \end{aligned}$

Step 6: solve for x by substituting $y=-\frac{208}{113}$ and $z = \frac{21}{113}$ into the first equation:

$2x+4(-\frac{208}{113})+32(\frac{21}{113})=6 \\ \\ 2x-\frac{832}{113}+\frac{672}{113}=6 \\ \\ 2x=6+\frac{832}{113}-\frac{672}{113} \\ \\ 2x=\frac{838}{113} \\ \\ x=\frac{319}{113}$

Finally:

$x = \frac{ 419 }{ 113 } ~,~y = -\frac{ 208 }{ 113 } ~,~z = \frac{ 21 }{ 113 }$

Learn more:

Solving System of Equations: brainly.com/question/13121177

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