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Solnce55 [7]
4 years ago
7

what is the value of the x variable in the solution to the following system of equations? 2x 2y = 6 x 3y = −1 a. −2 b. 2 c. 5 d.

−5
Mathematics
1 answer:
tatuchka [14]4 years ago
7 0
2x + 2y = 6 . . . . . (1)
x + 3y = -1 . . . . . (2)

(1) x 3 => 6x + 6y = 18 . . . . . (3)
(2) x 2 => 2x + 6y = -2 . . . . . (4)

(3) - (4) => 4x = 20
x = 20/4 = 5
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Natali5045456 [20]

Answer:

-2, -4, -3 + 2i, -3-2i

Step-by-step explanation:

Equaling the function to zero we have:

(x ^ 2 + 6x + 8) (x ^ 2 + 6x + 13) = 0  

For the first parenthesis we have:

(x ^ 2 + 6x + 8) = 0\\(x + 4) (x + 2) = 0  

Therefore the roots are:

x = - 4\\x = - 2  

For the second parenthesis we have:

(x ^ 2 + 6x + 13) = 0  

By completing squares we have:

x ^ 2 + 6x = -13  

x ^ 2 + 6x + (\frac{6}{2}) ^ 2 = -13 + (\frac{6}{2}) ^ 2\\x ^ 2 + 6x + 9 = -13 + 9\\(x + 3) ^ 2 = - 4\\x + 3 = +/- \sqrt{-4}    

Therefore the roots are:

x = -3 + 2i\\x = -3 - 2i  

Hope this was helpful

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3 years ago
I really don’t understand this and just need an explanation of how to do it.
sertanlavr [38]

\bf ~\hspace{12em}\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\cfrac{2n}{-3n\cdot -2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{-2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n\cdot -n}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n}\cdot \cfrac{1}{-n}


\bf \cfrac{1}{-3n}\cdot \boxed{1}\cdot \cfrac{1}{-n}\implies \cfrac{1}{-3n\cdot -n}\implies \cfrac{1}{3n^2}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4\implies \left( \cfrac{1}{3n^2} \right)^4\implies \stackrel{\textit{distributing the exponent}}{\cfrac{1^4}{3^4n^{2\cdot 4}}}\implies \cfrac{1}{81n^8}

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The equation can be made by:

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