Final answer:
Options B, C, and D represent systems of linear equations with infinitely many solutions because the second equation in each system can be derived by multiplying the coefficients and constants of the first equation by a common factor, signifying that both equations in the systems are equivalent and represent the same line.
Step-by-step explanation:
To find a system of 2 linear equations with infinitely many solutions, we need the two equations to be equivalent, which means they represent the same line on a graph. This can be determined by checking if the coefficients of x and y, as well as the constants, are in proportion in both equations.
Let's evaluate the given options:
- A) 2x + 3y = 9
4x + 6y = 18
Option A does not represent infinitely many solutions as the constants are not in the same proportion as the coefficients. The constants 9 and 18 are in proportion (9*2=18), but when we look at the coefficients, multiplying 2 and 3 by any number should result in 4 and 6 to have equivalent equations, and there's no single number that does that.
- B) 3x - 4y = 12
6x - 8y = 24
Option B shows two equations that are proportional. Multiplying the coefficients and constant of the first equation by 2 results in the coefficients and constant of the second equation (3*2=6, -4*2=-8, and 12*2=24), which suggests that these two equations are equivalent and represent the same line, hence they have infinitely many solutions.
- C) 5x + 6y = 18
15x + 18y = 54
In option C, all terms of the first equation are multiplied by 3 to obtain the second equation (5*3=15, 6*3=18, 18*3=54), indicating that they are equivalent and have infinitely many solutions.
- D) 10x - 15y = 45
2x - 3y = 9
In option D, the coefficients of the second equation can be multiplied by 5 to obtain the coefficients of the first equation (2*5=10 and -3*5=-15). Also, the constant 9 can be multiplied by 5 to obtain the constant 45 in the first equation (9*5=45), showing that these equations are also equivalent and represent the same line.
Hence, options B, C, and D have systems with infinitely many solutions, while option A does not.