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4 years ago

Find the ordered triple of these equations. 2x + 5y - 3z = -5 3x + 2y + 7z = 15 5x - 4y + 6z = 34

Mathematics

1 answer:

svetoff [14.1K]4 years ago

3 0

<h2>Ordered triple are (4,-2,1)</h2>

Step-by-step explanation:

We need to find ordered triple of 2x + 5y - 3z = -5 , 3x + 2y + 7z = 15 and 5x - 4y + 6z = 34

2x + 5y - 3z = -5 -------------------------eqn 1

3x + 2y + 7z = 15 -------------------------eqn 2

5x - 4y + 6z = 34 -------------------------eqn 3

eqn 1 + eqn 2

5x + 7y + 4z = 10 -------------------------eqn 4

eqn 4 - eqn 3

11y - 2z = -24 -------------------------eqn 5

eqn 1 x 3

6x + 15y - 9z = -15 -------------------------eqn 6

eqn 2 x 2

6x + 4y + 14z = 30 -------------------------eqn 7

eqn 7 - eqn 6

-11y + 23z = 45 -------------------------eqn 8

eqn 5 + eqn 8

21 z = 21

z = 1

Substituting in eqn 8

-11y + 23 x 1 = 45

-11 y = 22

y = -2

Substituting in eqn 1

2x + 5 x -2 - 3 x 1 = -5

2x = 8

x = 4

Ordered triple are (4,-2,1)

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<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%20%20%5Csqrt%7Bx%7D%20%20%2B%20%20%5Csqrt%5B5%5D%7Bx%7D%20" id="

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$3 + \frac{1}{2\sqrt{3} } + \frac{1}{5\sqrt[5]{81} }$

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Just to make it easier to see, $\sqrt{x} = x^{\frac{1}{2} }$ and $\sqrt[5]{x} = x^{\frac{1}{5} }$ This way we could more easily use the power rule of derivatives.

So if f(x) = $x^{2} +x^{\frac{1}{2} } +x^{\frac{1}{5} }$ then f'(x) will be as follows.

f'(x) = $x^{1} +\frac{1}{2} x^{-\frac{1}{2} } +\frac{1}{5} x^{-\frac{4}{5} } = x +\frac{1}{2x^{\frac{1}{2} }} +\frac{1}{ 5x^{\frac{4}{5} }} = x +\frac{1}{2\sqrt{x}} +\frac{1}{ 5\sqrt[5]{x^4} }$

to find f'(3) just plug 3 into f'(x) so $3 + \frac{1}{2\sqrt{3} } + \frac{1}{5\sqrt[5]{81} }$

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