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14 September, 12:32

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

3x - 2y = 3; 6x - 4y = 1

3x - 5y = 8,5x - - 3y = 2

3x + 2y = 8; 4x + 3y = 1

3x - y = 3; 2x - 4y = 2

3x - 4y = 2; 6x - y = 1

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  1. 14 September, 15:37
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    First equation: no solution

    Second equation: one solution

    Third equation: one solution

    Fourth equation: one solution

    Fifth equation: one solution

    Step-by-step explanation:

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

    Multiplying the first equation by two, we have:

    6x - 4y = 6

    As this equation and the second equation of the system have the same coefficients to x and y, but a different independent value (1 and 6), the 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, we have:

    (40 + 25y) / 3 - 3y = 2

    40 + 25y - 9y = 6

    16y = 34

    y = 2.125

    x = (8 + 5*2.125) / 3 = 6.2083

    The 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, we have:

    (32 - 8y) / 3 + 3y = 1

    32 - 8y + 9y = 3

    y = - 29

    x = (8 - 2 * (-29)) / 3 = 22

    The system has one solution

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

    From the first equation: y = 3x - 3

    Using this value of y in the second equation, we have:

    2x - 12x + 12 = 2

    10x = 10

    x = 1

    y = 3 - 3 = 0

    The system has one solution

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

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

    Using this value of 3x in the second equation, we have:

    4 + 8y - y = 1

    7y = - 3

    y = - 0.4286

    x = (2 + 4y) / 3 = (2 + 4 * (-0.4286)) / 3 = 0.0952

    The system has one solution
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