Question

A system of linear equations is shown below, where A and B are real numbers.

3x + 4y = A

Bx – 6y = 15

What values could A and B be for this system to have no solutions?

Answers

  1. Answer:
    A = 0; B = -9/2
    Step-by-step explanation:
    To have no solutions, you need parallel lines with equal slopes and different y-intercepts.
    3x + 4y = A      Eq. 1
    Bx – 6y = 15     Eq. 2
    In Eq. 1, notice that the coefficient of x is 3/4 of the coefficient of y.
    We must have the same ratio for the coefficients in Eq. 2.
    B/(-6) = 3/4
    4B = -6(3)
    4B = -18
    B = -9/2
    Now we have
    3x + 4y = A      Eq. 1
    -9/2 x – 6y = 15     Eq. 2
    How do we change the left side of the second equation into the left side of the first equation? -6/4 = -3/2 and also -9/2 ÷ 3 = -3/2
    To change the left side of the second equation into the left side of the first equation, divide the left side by -3/2.
    If we divide 15 by -3/2 we get -10.
    The equation -9/2 x – 6y = -10 is the same as Eq. 1, so that would create a system of equations with only one equation and an infinite number of answers.
    To have no equations, the y-intercepts must be different, so A can be any number other that -10.
    Answer: A = 0; B = -9/2

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