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YOU MUST USE ALL 6 EQUATIONS from "a" to answer c,d, and e. You cannot use an equation more than once. c. What two equations from those above would form a system with infinite solutions. You can list the 2 equations here: and d. What two equations from those above would form a system with no solution. You can list the 2 equations here: and e. What two equations from those above would form a system with one solution. AND give the solution (x,y) for both equations (x, y) HERE: ( ) You can list the 2 equations here: and MUST SHOW WORK ON HOW YOU GOT YOUR SOLUTION

1 answer

To address your question, we'll need to first know the actual equations labeled "a." Without seeing them, I can only guide you on the general concepts related to systems of equations.

c. Infinite Solutions

Two equations will have infinite solutions if they are the same line, meaning one equation is simply a multiple of the other. For example, if one equation can be represented as a scalar multiple of another, they will intersect at every point along that line.

To find such pairs, check if one can be transformed into the other through multiplication or division.

d. No Solution

Two equations will have no solution if they are parallel lines; that is, they have the same slope but different y-intercepts. To determine if two equations are parallel, compare their slopes.

e. One Solution

Two equations will have one solution if their lines intersect at exactly one point. This occurs when they have different slopes.

To find the intersection point, you can solve the two equations either by substitution or elimination.

Example

Assuming the 6 equations are:

  1. \(y = 2x + 3\)
  2. \(y = 4x - 1\)
  3. \(2y = 4x + 6\) (which becomes the same as equation 1)
  4. \(y = 0.5x + 1\)
  5. \(2y = 2x + 2\) (which simplifies to \(y = x + 1\))
  6. \(y = -x + 2\)

Using these equations:

c. Infinite Solutions:

  • Equation 1: \(y = 2x + 3\)
  • Equation 3: \(2y = 4x + 6\) (this simplifies to the same equation)

d. No Solution:

  • Equation 1: \(y = 2x + 3\)
  • Equation 2: \(y = 4x - 1\) (these have different slopes)

e. One Solution:

  • Equation 2: \(y = 4x - 1\)
  • Equation 6: \(y = -x + 2\)

To find the intersection point: Set the two equations equal: \[4x - 1 = -x + 2\]

Now, solve for \(x\): \[ 4x + x = 2 + 1 \] \[ 5x = 3 \] \[ x = \frac{3}{5} \]

Substitute \(x\) back into one of the equations to find \(y\): Using \(y = 4x - 1\): \[ y = 4\left(\frac{3}{5}\right) - 1 = \frac{12}{5} - 1 = \frac{12}{5} - \frac{5}{5} = \frac{7}{5} \]

Thus, the solution (x,y) is:

(x,y) = \left(\frac{3}{5}, \frac{7}{5}\right)

Final Answers

c. The two equations are:

  • \(y = 2x + 3\)
  • \(2y = 4x + 6\)

d. The two equations are:

  • \(y = 2x + 3\)
  • \(y = 4x - 1\)

e. The two equations are:

  • \(y = 4x - 1\)
  • \(y = -x + 2\)

Solution is:

  • (x,y) = \(\left(\frac{3}{5}, \frac{7}{5}\right)\)

Make sure to adjust this appropriately according to the specific equations you are working with!