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Sphinxa [80]
3 years ago
7

Solve the following systems of 3-variable linear equations.​50 points

Mathematics
1 answer:
Natali [406]3 years ago
7 0

Answer:

x = 9/25

y = 7/25

z = 4/25

Step-by-step explanation:

2x + y = 1 .......(1)

3y + z = 1 ........(2)

x + 4z = 1 ........(3)

Elimination 1 and 2

2x + y = 1 | ×3 |

3y + z = 1 | ×1 |

6x + 3y = 3

3y + z = 1

___________--

6x - z = 2 .............. (4)

Elimination 3 and 4

x + 4z = 1 | ×6 |

6x - z = 2 | ×1 |

6x + 24z = 6

6x - z = 2

___________--

25z = 4

z = 4/25

Elimination 3 and 4

x + 4z = 1 | ×1 |

6x - z = 2 | ×4 |

x + 4z = 1

24x - 4z = 8

___________+

25x = 9

x = 9/25

Subsitution 1

2x + y = 1

2(9/25) + y = 1

18/25 + y = 1

y = 1 - 18/25

y = 25/25 - 18/25

y = 7/25

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nlexa [21]

From the given table, we have that the lateral limits of f(x) as x -> 3 are different, hence the limit of f(x) does not exist at x = 3.

<h3>What is a limit?</h3>

A limit is given by the value of function f(x) as x tends to a value. For the limit to exist, the lateral limits have to be the same, as follows:

\lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x)

In this problem, we have that:

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Hence the lateral limits are given as follows:

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Since the lateral limits are different, the limit does not exist.

More can be learned about lateral limits at brainly.com/question/26270080

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