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4 March, 21:36

Choose all of the system of equations which have no solution. x + 2y = 10 and 2y = 6 + 5x x = 8 - y and y = 6 + 5x 3x + 2y = 5 and 2y = 6 - 3x 3x = 2 - 4y and 4y = 7 - 3x

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  1. 4 March, 23:11
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    Third and fourth systems

    Step-by-step explanation:

    Let's check the first system:

    x + 2y = 10

    2y = 6 + 5x

    Isolating x in the first equation, we have x = 10 - 2y. Applying that in the second equation we have:

    2y = 6 + 5 (10-2y)

    2y = 6 + 50 - 10y

    12y = 56

    y = 4.6667

    from the first equation, we have that x + 2*4.6667 = 10 - > x = 0.6667

    So this system has a solution.

    Checking now the second system:

    x = 8 - y

    y = 6 + 5x

    using y from the second equation in the first equation, we have:

    x = 8 - 6 - 5x

    6x = 2

    x = 0.3333

    then, in the second equation:

    y = 6 + 5*0.3333 = 7.6666

    This system also has a solution

    Third system:

    3x + 2y = 5

    2y = 6 - 3x

    The second equation can be rewritten as:

    3x + 2y = 6

    We can see from both equation that the same expression (3x + 2y) has two values (5 and 6). So we can't solve this system.

    Fourth system:

    3x = 2 - 4y

    4y = 7 - 3x

    The first equation can be rewritten as 3x + 4y = 2

    The second equation can be rewritten as 3x + 4y = 7

    We can see from both equation that the same expression (3x + 4y) has two values (2 and 7). So we can't solve this system.

    So, the systems that have no solution are the third and fourth systems
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