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  • Brainly User
2015-10-24T11:47:47+08:00

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This is quadratic, not linear, equation.

x+3  =    x-3  
x+ 4     5x - 3

(x+3) (5x-3) = (x+4) (x-3)
5x² - 3x + 15x - 9 = x² - 3x + 4x - 12
5x² + 12x - 9 = x² + x -12

Transform to ax² + bx + c = 0
5x² - x² + 12x - x - 9 + 12 = 0
4x² + 11x + 3 = 0

Solve using quadratic formula (because this equation can not be factored with rational numbers).
4x² + 11x + 3 = 0
a = 4;  b = 11;  c = 3

x =  \frac{-b+or- \sqrt{b ^{2}+4ac } }{2a}

x =  \frac{-11+or- \sqrt{(11) ^{2}-4(4)(3) } }{2(4)}

x =  \frac{-11+or- \sqrt{121-48} }{8}

x =  \frac{-11+or- \sqrt{73} }{8}

The roots are irrational:

x =   \frac{-11+ \sqrt{73} }{8}
         and
x= \frac{-11- \sqrt{73} }{8}

(Or you may refer to may handwritten solution.  Please see attached.)




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