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OverLord2011 [107]
2 years ago
8

Can you please help me with this problem?

Mathematics
1 answer:
Hoochie [10]2 years ago
6 0
Here you have to find which each variable is, for this you start of picking one equation,

x + 2y + 6z = 4
-3x + 2y - 2 = -4
4x + 2z = 16

depending the equation you pick you multiply that by a certain number that will give you the opposite of one of the other equations,
-1(x + 2y + 6z = 4)
= -x -2y - 6z = -4

With this you add or subtract it with the equation that has the same number or variable, or both,
In this case it will be the equation,
-3x + 2y + 6z = 4
You can use this one or the third equation since both have a positive 2y which will cancel with -2y from the new equation,
-x - 2y - 6z = -4
-3x + 2y -z = -4

= -4x -7z = -8
Now you since you just eliminated the variable (y) you now have 2 variables, and the last equation has only 2 variables, meaning now you find the answer to those to equations,

-4x -7z = -8
4x + 2z = 16

= -5z = 8
Now leave the variable by itself,
z = 8/5
Now you found the variable (z), with this just substitute on one of the equations we used to find (z) so you can find (x), after that substitute those answered to on of the original equations so you can find (y)




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Answer:

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Step-by-step explanation:

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^^

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3 years ago
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rjkz [21]
<h3>2 Answers: Choice B, Choice C</h3>

The rule we use here is \frac{a^b}{a^c} = a^{b-c}

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Answer:

Third graph C.

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Read 2 more answers
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