how many solutions does each system have?

system: 1
4x+3y=-1
12x+9y=-21
system 2
y=2x+3
4x-2y=-6
Both systems have an infinite number of solutions.


Both systems have no solution.


System I has an infinite number of solutions and System II has no solution.


System I has no solution and System II has an infinite number of solutions.

1 answer

To analyze both systems and determine the number of solutions for each, we need to check the relationships between the equations in each system.

System 1:

  1. \( 4x + 3y = -1 \)
  2. \( 12x + 9y = -21 \)

First, notice that the second equation can be simplified. We can divide the entire equation by 3:

\[ \frac{12x + 9y}{3} = \frac{-21}{3} \quad \Rightarrow \quad 4x + 3y = -7 \]

Now we have the system:

  1. \( 4x + 3y = -1 \)
  2. \( 4x + 3y = -7 \)

These two equations clearly describe parallel lines (same left-hand side, different right-hand sides), which means they will never intersect. Therefore, System 1 has no solution.

System 2:

  1. \( y = 2x + 3 \)
  2. \( 4x - 2y = -6 \)

Let's substitute \( y \) from the first equation into the second equation. Replacing \( y \) with \( 2x + 3 \):

\[ 4x - 2(2x + 3) = -6 \]

Distributing the -2 yields:

\[ 4x - 4x - 6 = -6 \]

This simplifies to:

\[ -6 = -6 \]

This is always true, indicating that the two equations describe the same line. Therefore, System 2 has an infinite number of solutions.

Conclusion:

  • System I has no solution.
  • System II has an infinite number of solutions.

Thus, the correct statement is: System I has no solution and System II has an infinite number of solutions.