The problem is x/x-5 (first FRACTION) +4 = 1/x+3 (2nd FRACTION) Is x=4 a solution
F FMMurphy Junior Member Joined Mar 12, 2006 Messages 51 May 16, 2007 #1 The problem is x/x-5 (first FRACTION) +4 = 1/x+3 (2nd FRACTION) Is x=4 a solution
J jwpaine Full Member Joined Mar 10, 2007 Messages 723 May 17, 2007 #2 \(\displaystyle \L\frac{x}{x-5} + \frac{4}{1} = \frac{1}{x+3}\) To add fractions you need a common denominator...so: \(\displaystyle \L\frac{x(1)}{x-5} + \frac{4(x-5)}{(x-5)} = \frac{1}{x+3}\) \(\displaystyle \L\frac{x + 4x-20}{x-5} = \frac{1}{x+3}\) \(\displaystyle \L\frac{5x-20}{x-5} = \frac{1}{x+3}\) Cross multiply: \(\displaystyle \L 1(x-5) = (x+3)(5x-20)\) FOIL the right side. \(\displaystyle \L x-5 = 5x^2-5x-60\) Combine both sides (set the quadratic equal to zero). \(\displaystyle \L 5x^2-6x-55 = 0\) Now solve your new polynomial for it's two solutions...... (you can also plug in x = 4 into f(x) = 5x^2-6x-55 to see if it's a zero) Best of luck!
\(\displaystyle \L\frac{x}{x-5} + \frac{4}{1} = \frac{1}{x+3}\) To add fractions you need a common denominator...so: \(\displaystyle \L\frac{x(1)}{x-5} + \frac{4(x-5)}{(x-5)} = \frac{1}{x+3}\) \(\displaystyle \L\frac{x + 4x-20}{x-5} = \frac{1}{x+3}\) \(\displaystyle \L\frac{5x-20}{x-5} = \frac{1}{x+3}\) Cross multiply: \(\displaystyle \L 1(x-5) = (x+3)(5x-20)\) FOIL the right side. \(\displaystyle \L x-5 = 5x^2-5x-60\) Combine both sides (set the quadratic equal to zero). \(\displaystyle \L 5x^2-6x-55 = 0\) Now solve your new polynomial for it's two solutions...... (you can also plug in x = 4 into f(x) = 5x^2-6x-55 to see if it's a zero) Best of luck!