To solve the given systems of equations, we will analyze each system and determine the number of solutions for each.
System 1:
- \( 2x - 2y = -4 \)
- \( y = x + 2 \)
First, let's rearrange the first equation to solve for \(y\): \[ 2x - 2y = -4 \] \[ -2y = -4 - 2x \] \[ 2y = 4 + 2x \] \[ y = 2 + x \]
Now, comparing the two equations:
- From the first equation rearranged: \( y = x + 2 \)
- From the second equation: \( y = x + 2 \)
Since both lines are the same, there are infinitely many solutions.
System 2:
- \( y = 3x - 2 \)
- \( 2x + 3y = 16 \)
Now we can substitute the first equation into the second:
- Substitute \( y \) in the second equation: \[ 2x + 3(3x - 2) = 16 \] \[ 2x + 9x - 6 = 16 \] \[ 11x - 6 = 16 \] \[ 11x = 22 \] \[ x = 2 \]
Now substitute back to find \(y\): \[ y = 3(2) - 2 = 6 - 2 = 4 \]
Thus, there is one solution at the point \( (2, 4) \).
System 3:
- \( y = -2x - 7 \)
- \( y = -2x + 3 \)
Comparing the two equations:
- From the first equation: \( y = -2x - 7 \)
- From the second equation: \( y = -2x + 3 \)
Since the slopes (-2) are equal but the y-intercepts are different (-7 and +3), these lines are parallel. Thus, there is no solution.
Summary:
- System 1: Infinitely many solutions
- System 2: One solution
- System 3: No solution
Matching Systems to Number of Solutions:
- Infinitely many solutions: for \(2x - 2y = -4\) and \(y = x + 2\)
- One solution: for \(y = 3x - 2\) and \(2x + 3y = 16\)
- No solution: for \(y = -2x - 7\) and \(y = -2x + 3\)
Please match the systems accordingly in your input.