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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?

User Gurnzbot
by
8.3k points

1 Answer

6 votes

Answer:

A = 0; B = -9/2

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

User Synexis
by
8.1k points

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