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Identify any two equations that form a system with one solution (different slopes) ? **Hint: You must write # 4 - #6 in y = mx + b format first.

1. y = 5x - 4
2. y = -2 x + 3
3. y = 3x + 1
4. 2y = 6x + 2
5. y - 5x = 2
6. y - 4x + 1 = 0

1 answer

To identify two equations from the provided list that form a system with one solution, we need to express each equation in the slope-intercept form (y = mx + b). We will work on equations 4, 5, and 6.

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

    • To convert it to the slope-intercept form, divide both sides by 2: \[ y = 3x + 1 \] (This equation is already present as equation 3.)

  2. Equation 5: \(y - 5x = 2\)

    • Rearranging this to slope-intercept form gives: \[ y = 5x + 2 \]

  3. Equation 6: \(y - 4x + 1 = 0\)

    • Rearranging this to slope-intercept form gives: \[ y = 4x - 1 \]

Now we have:

  • Equation 3: \(y = 3x + 1\)
  • Equation 5: \(y = 5x + 2\)
  • Equation 6: \(y = 4x - 1\)

Next, let's check the slopes of the pairs:

  • Slope of Equation 3: \(m_1 = 3\)
  • Slope of Equation 5: \(m_2 = 5\)
  • Slope of Equation 6: \(m_3 = 4\)

The combinations with different slopes are:

  • Equations 5 and 6: \(y = 5x + 2\) and \(y = 4x - 1\) (Their slopes are 5 and 4 respectively, so they intersect at one point.)
  • Equations 3 and 5: \(y = 3x + 1\) and \(y = 5x + 2\) (Different slopes, 3 and 5.)
  • Equations 3 and 6: \(y = 3x + 1\) and \(y = 4x - 1\) (Different slopes, 3 and 4.)

Therefore, any of the following pairs can be used to form a system with one solution:

  • Equations 5 and 6: \(y = 5x + 2\) and \(y = 4x - 1\)
  • Equations 3 and 5: \(y = 3x + 1\) and \(y = 5x + 2\)
  • Equations 3 and 6: \(y = 3x + 1\) and \(y = 4x - 1\)

Example Pair: \(y = 5x + 2\) and \(y = 4x - 1\) (Equations 5 and 6)