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Shtirlitz [24]
3 years ago
10

Determine whether the systems have one solution, no solution, or infinitely many solutions

Mathematics
1 answer:
Lerok [7]3 years ago
5 0

Answer:

First system: no solution

Second system: one solution

Third system: one solution

Fourth system: infinite solutions

Fifth system: no solution

Step-by-step explanation:

First system: 3x-2y=3; 6x-4y=1

From the first equation: y = (3x - 3)/2

Using this value of y in the second equation:

6x - 6x + 6 = 1

6 = 1 -> System has no solution

Second system: 3x-5y=8; 5x-3y=2

From the first equation: x = (8 + 5y)/3

Using this value of x in the second equation:

5*(8 + 5y) - 9y = 6

40 + 25y - 9y = 6

16y = -34 -> y = -2.125

x = (8 - 5*2.125)/3 = -0.875

This system has one solution

Third system: 3x-2y=8; 4x-3y=1

 From the first equation: x = (8 + 2y)/3

Using this value of x in the second equation:

4*(8 + 2y) - 9y = 3

32 + 8y - 9y = 6

y = 26

x = (8 + 2*26)/3 = 20

This system has one solution

Fourth system: 3x-6y=3; 2x-4y=2

  From the first equation: x = 1 + 2y

Using this value of x in the second equation:

2*(1 + 2y) - 4y = 2

2 + 4y - 4y = 2

2 = 2

This system has infinite solutions

Fifth system: 3x-4y=2; 6x-8y=1

  From the first equation: x = (2 + 4y)/3

Using this value of x in the second equation:

2*(2 + 4y) - 8y = 1

4 + 8y - 8y = 2

4 = 2

This system has no solution

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Answer to this question ?
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The answer is maybe a or b
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