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
[tex]x = \frac{-b+or- \sqrt{b ^{2}+4ac } }{2a} [/tex]
[tex]x = \frac{-11+or- \sqrt{(11) ^{2}-4(4)(3) } }{2(4)} [/tex]
[tex]x = \frac{-11+or- \sqrt{121-48} }{8} [/tex]
[tex]x = \frac{-11+or- \sqrt{73} }{8} [/tex]
The roots are irrational:
[tex]x = \frac{-11+ \sqrt{73} }{8} [/tex]
and
[tex]x= \frac{-11- \sqrt{73} }{8} [/tex]
(Or you may refer to may handwritten solution. Please see attached.)
