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larisa [96]
3 years ago
5

Does anyone know this

Mathematics
1 answer:
krek1111 [17]3 years ago
4 0

Answer:

<em>a2+1)-(4+2a2) </em>

Step-by-step explanation:

<em>to solve this, we need to get rid of the brackets as follow: </em>

<em>3a2+1-4-2a2  </em>

<em>ading the common terms we have: </em>

<em>a2 - 3 </em>

<em>that is all.</em>

<em>Plz mark  me brainlist!!!</em>

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Step-by-step explanation:

For example, the most common way to write a ratio is as a fraction, 3/6. We could also write it using the word "to," as "3 to 6." Finally, we could write this ratio using a colon between the two numbers, 3:6. Be sure you understand that these are all ways to write the same number.

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Solve the system of equations.
emmainna [20.7K]

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

<h2>Explanation:</h2>

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 }

<h2>Learn more:</h2>

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

#LearnWithBrainly

7 0
4 years ago
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