ANSWER
The equation has no solution if
\(A=6\) and \(B=-4.5\).
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;
\(y=\frac{AB-45}{4B+18}\)
The above equation is defined for all real values except
\(4B+18=0\)
This implies that
\(B=-\frac{18}{4}\)
\(B=-4.5\)
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
\(-4.5A
e 45\)
\(\Rightarrow A
e -\frac{45}{4.5}\)
\(\Rightarrow A
e -10\)
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.
