Question 464264: I would sincerely appreciate it if you could please help me solve the following equation:
23/(x-1) + 3 - 70/(x^2-4x+3)=0
I know that when x^2-4x+3 is factored, it becomes (x-1)(x-3), and I think I need to get rid of the denominator to solve the question, but I'm experiencing some difficulty finishing the question.
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! 23/(x-1) + 3 - 70/(x^2-4x+3) = 0
23/(x - 1) + 3 - 70/(x - 3)(x - 1) = 0 {factored x² - 4x + 3 into two binomials}
(x - 3)(x - 1)[23/(x - 1) + 3 - 70/(x - 3)(x - 1) = 0] {multiplied entire equation by LCD, (x - 3)(x - 1)}
Multiply each term by (x - 3)(x - 1):
(x - 3)(x - 1)[23/(x - 1)] + (x - 3)(x - 1)(3) - (x - 3)(x - 1)[70/(x - 3)(x - 1) = (x - 3)(x - 1)(0)
23(x - 3) + 3(x² - 4x + 3) - 70 = 0 {cancelled and multiplied what's left}
23x - 69 + 3x² - 12x + 9 - 70 = 0 {used distributive property}
3x² + 11x - 130 = 0 {combined like terms}
(3x + 26)(x - 5) = 0 {factored into two binomials}
3x + 26 = 0 or x - 5 = 0 {set each factor equal to 0}
3x = -26 or x = 5 {subtracted 26 and added 5, respectively}
x = -26/3 or x = 5 {divided first equation by 3}
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