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?

A. A = 6, B = –4.5
B .A = –10, B = –4.5
C. A = –6, B = –3
D. A = 10, B = –3

Answer 1

Answer: A

Step-by-step explanation:just in case you didn't wanna read all that

Answer 2

The equation has no solution if

`A=6` and
`B=-4.5`.

Step-by-step explanation

The first equation is

`3x+4y=A--(1)`

and the second equation is

`Bx-6y=15--(2)`

We try solving this equation for B using the method of elimination.

We multiply equation (2) by 3 to obtain;

`3Bx-18y=45--(3)`

We also multiply equation (1) by B to obtain;

`3Bx+4By=AB---(4)`

We now subtract equation (3) from equation (4).

This will give us;

`4By+18y=AB-45`

We factor
`y` on the right hand side to get;

`(4B+18)y=AB-45`

We solve for y to obtain;

`y=(AB-45)/(4B+18)`

The above equation is defined for all real values except

`4B+18=0`

This implies that

`B=-(18)/(4)`

`B=-4.5`

Remember that the system will have no solution if and only if the denominator is zero and the numerator is not equal to zero.

This means that,

`AB-45\\e 0`

`\Rightarrow AB\\e 45`

When we substitute
`B=-4.5`, then we will have

`-4.5A\\e 45`

`\Rightarrow A\\e -(45)/(4.5)`

`\Rightarrow A\\e -10`

Hence the correct answer is option A

Note however that if both the numerator and denominator equal zero, then the system has infinitely many solutions.