Which of the following systems of linear equations has an infinite number of solutions?

Here we do not have the options, but ill try to give an answer.

A system of equations is something of the shape:

A1*x + B1*y = C1

A2*x + B2*y = C2

Where this is a linear relationship, you could have a more complex system, but this will work for the example.

If you want to have a unique solution for a system of equations, you need to have the same number of linearly independent equations (this means that the equations are not repeated, like for example in x + y = 2, and 2x + 2y = 4, where the second equation is equal to two times the first one) than the number of variables, so if we have two variables X and Y, we need to have two linearly independent equations.

If we have less, suppose that you have two variables and only one equation, you will have:

A1*x + B1*y = C1

y = (C1 - A1*x)/B1

Here, for each value of x you have a different value of y, so you have infinite solutions.

So for an answer this you need to find the system where the number of linearly independent equations is smaller than the number of variables.

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18