Question

which system of equations has a solution of approximately (-0.6, 0.8)?
○ 2x - 5y = -5 and 14x + 5y = -5
○ 2x - 5y = -5 and 11x - 5y = 15
○ 11x - 5y = 15 and 14x + 5y = -5
○ 11x - 5y = 15 and 9x + 5y = 5

Explanation:

Step1: Test the point (-0.6, 0.8) in the first system

For the system \(2x - 5y = -5\) and \(14x + 5y = -5\):

• Substitute \(x = -0.6\), \(y = 0.8\) into \(2x - 5y\):

\(2(-0.6)-5(0.8)= -1.2 - 4 = -5.2\approx -5\) (close)

• Substitute into \(14x + 5y\):

\(14(-0.6)+5(0.8)= -8.4 + 4 = -4.4\approx -5\) (close)

Step2: Test the point in the second system

For \(2x - 5y = -5\) and \(11x - 5y = 15\):

• Substitute into \(11x - 5y\):

\(11(-0.6)-5(0.8)= -6.6 - 4 = -10.6
eq15\)

Step3: Test the point in the third system

For \(11x - 5y = 15\) and \(14x + 5y = -5\):

• Substitute into \(11x - 5y\):

\(11(-0.6)-5(0.8)= -6.6 - 4 = -10.6
eq15\)

Step4: Test the point in the fourth system

For \(11x - 5y = 15\) and \(9x + 5y = 5\):

• Substitute into \(11x - 5y\):

\(11(-0.6)-5(0.8)= -6.6 - 4 = -10.6
eq15\)

• Substitute into \(9x + 5y\):

\(9(-0.6)+5(0.8)= -5.4 + 4 = -1.4
eq5\)

Answer:

A. \(2x - 5y = -5\) and \(14x + 5y = -5\)