Which system of equations has infinitely many solutions?

Question

4 x + 2 y = 5. Negative 4 x minus 2 y = 1.
Negative 10 x + y = 4. 10 x minus y = negative 4.
Negative 8 x + y = 2. 8 x minus y = 0.
Negative x + 2 y = 6. 7 x minus 2 y = 12.

Answer 1

To determine which system of equations has infinitely many solutions, we need to check if any of the equations in a pair are equivalent (i.e., one can be derived from another by multiplying the entire equation by a non-zero constant).

Let's analyze each option:

1. First set of equations: \[ 4x + 2y = 5 \] \[ -4x - 2y = 1 \] These equations are not multiples of each other. In fact, they represent two distinct lines that will intersect at one point (thus one solution) since they have different slopes.

2. Second set of equations: \[ -10x + y = 4 \] \[ 10x - y = -4 \] Rearranging the second equation gives \(10x - y + 4 = 0\) or \(y = 10x + 4\). These equations have different slopes and represent different lines, so they will also have only one solution.

3. Third set of equations: \[ -8x + y = 2 \] \[ 8x - y = 0 \] The second equation can be simplified to \(y = 8x\). If we substitute \(y = 8x\) into the first equation, we get: \[ -8x + 8x = 2 \] This simplifies to \(0 = 2\), which is a contradiction, indicating that these two lines are parallel and do not intersect (no solution).

4. Fourth set of equations: \[ -x + 2y = 6 \] \[ 7x - 2y = 12 \] The first equation can be rewritten as \(2y = x + 6\) or \(y = \frac{1}{2}x + 3\). The second equation can be manipulated to find a relationship, but it is not a multiple of the first equation, which means they represent two lines that intersect at one point (one solution).

From our analysis, the set that might possibly yield infinitely many solutions is the one where two lines are equivalent. Examining this closely, it becomes clear that none of the above settings have equivalent lines; rather, they all produce either a single point of intersection or are parallel (no solution).

Conclusion: Upon reviewing all sets, none of the equations in these systems yield infinitely many solutions. However, if they were equivalent in pairings or had matching forms through modifications, they would present infinitely many solutions. Thus, under the provided four choices, none meet the criteria for infinitely many solutions.