a system of linear equations is shown below, where a and b are real numbers.\n3x + 4y = a\nbx - 6y =…

a system of linear equations is shown below, where a and b are real numbers.\n3x + 4y = a\nbx - 6y = 15\nwhat values could a and b be for this system to have no solutions?\na = 6, b = -4.5\na = -10, b = -4.5\na = -6, b = -3\na = 10, b = -3

a system of linear equations is shown below, where a and b are real numbers.\n3x + 4y = a\nbx - 6y = 15\nwhat values could a and b be for this system to have no solutions?\na = 6, b = -4.5\na = -10, b = -4.5\na = -6, b = -3\na = 10, b = -3

Answer

Explanation:

Step1: Condition for no solutions

For a system of linear equations (a_1x + b_1y=c_1) and (a_2x + b_2y=c_2), the system has no solutions if (\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}). For the given system (3x + 4y=A) ((a_1 = 3), (b_1=4), (c_1 = A)) and (Bx-6y = 15) ((a_2=B), (b_2=-6), (c_2 = 15)). First, find (\frac{a_1}{a_2}=\frac{b_1}{b_2}), i.e., (\frac{3}{B}=\frac{4}{-6}). Cross - multiply: (3\times(-6)=4B). (- 18=4B), so (B=-\frac{9}{2}=-4.5).

Step2: Check (\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2})

When (B = - 4.5), check (\frac{3}{-4.5}=\frac{4}{-6}=-\frac{2}{3}). Now check (\frac{A}{15}\neq-\frac{2}{3}). Cross - multiply: (A\neq-\frac{2}{3}\times15), so (A\neq - 10).

Answer:

(A = 6,B=-4.5)