ANSWER
The equation has no solution if
[tex]A=6[/tex] and [tex]B=-4.5[/tex].
EXPLANATION
The first equation is
[tex]3x+4y=A--(1)[/tex]
and the second equation is
[tex]Bx-6y=15--(2)[/tex]
We try solving this equation for B using the method of elimination.
We multiply equation (2) by 3 to obtain;
[tex]3Bx-18y=45--(3)[/tex]
We also multiply equation (1) by B to obtain;
[tex]3Bx+4By=AB---(4)[/tex]
We now subtract equation (3) from equation (4).
This will give us;
[tex]4By+18y=AB-45[/tex]
We factor [tex]y[/tex] on the right hand side to get;
[tex](4B+18)y=AB-45[/tex]
We solve for y to obtain;
[tex]y=\frac{AB-45}{4B+18}[/tex]
The above equation is defined for all real values except
[tex]4B+18=0[/tex]
This implies that
[tex]B=-\frac{18}{4}[/tex]
[tex]B=-4.5[/tex]
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,
[tex]AB-45\ne 0[/tex]
[tex]\Rightarrow AB\ne 45[/tex]
When we substitute [tex]B=-4.5[/tex], then we will have
[tex]-4.5A\ne 45[/tex]
[tex]\Rightarrow A\ne -\frac{45}{4.5}[/tex]
[tex]\Rightarrow A\ne -10[/tex]
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.
