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?

1. 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