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help please, will give brainliest!
A. the equation has two true solutions
B. the equation had one true solution
C. the equation has one extraneous solution
D. the equation has one true solution and one extraneous solution ​

Help Please Will Give BrainliestA The Equation Has Two True Solutions B The Equation Had One True Solution C The Equation Has One Extraneous Solution D The Equa class=

Sagot :

mhanifa

Answer:

  • C. the equation has one extraneous solution

Step-by-step explanation:

Given equation:

  • x/(x - 1) + 2/(x + 2) = -6/(x² + x - 2)

Common denominator is x² + x - 2, consider x ≠ 1 and x≠ -2

Multiply both sides by x² + x - 2:

  • x(x + 2) + 2(x - 1) = -6
  • x² + 2x + 2x - 2 + 6 = 0
  • x² + 4x + 4 = 0
  • (x + 2)² = 0
  • x + 2 = 0
  • x = -2

There is one solution but it is not true one.

Option C is correct

manissaha129

Answer:

[tex] \frac{x}{x - 1} + \frac{2}{x + 2} = \frac{ - 6}{ {x}^{2} + x - 2 } \\ \frac{x(x + 2) + 2(x - 1)}{(x - 1)(x + 2)} = \frac{ - 6}{ {x}^{2} + x - 2} \\ \frac{ {x}^{2} + 2x + 2x - 2 }{ {x}^{2} + 2x - x - 2} = \frac{ - 6}{ {x}^{2} + x - 2} \\ \frac{ {x}^{2} + 4x - 2 }{ {x}^{2} + x - 2} = \frac{ - 6}{ {x}^{2} + x - 2 } \\ {x}^{2} + 4x - 2 = - 6 \\ {x}^{2} + 4x + 4 = 0 \\ {x}^{2} + 2 \times 2 \times x + {2}^{2} = 0 \\ {(x + 2)}^{2} = 0 \\ x + 2 = 0 \\ \boxed{x = - 2 }[/tex]

x =-2 is the right answer.

C. the equation has one extraneous solution.

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