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

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,

$AB-45\ne 0$

$\Rightarrow AB\ne 45$

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

$-4.5A\ne 45$

$\Rightarrow A\ne -\frac{45}{4.5}$

$\Rightarrow A\ne -10$

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.