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umka2103 [35]
3 years ago
12

One equation in a system of equations is given in the graph. Which equation below would have to be the second equation so that t

he system has no solutions?

Mathematics
1 answer:
Tems11 [23]3 years ago
6 0

Solution:

To classify the system of equations as no solutions, the lines must be parallel (Not intersect).

<u>First, let's identify the equation forming the line.</u>

  • Slope = Rise/Run = 2/-3 = -0.66
  • Y-intercept = 0
  • Equation formed: y = -2/3

We also know that the slope of the equation must be -1.5.

<u>A) 2x + 3y = 0</u>

  • 3y = 0 - 2x
  • => 3y = -2x
  • => (y = -2x/3) ∦ (y = -2x/3)

<u>B) 2x + 3y = 6</u>

  • => 3y = -2x + 6
  • => (y = -2x/3 + 2) ║ (y = -2x/3)

<u>C) 2x - 3y = 9</u>

  • => -3y = -2x + 9
  • => y = -2x/-3 + 9/-3
  • => (y = 2x/3 - 3) ∦ (y = -2x/3)

<u>D) y = 3x + 2</u>

  • (y = 3x + 2) ∦ (y = -2x/3)

Option B is correct.

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