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solutions to rational equation
- Thread starter banana
- Start date
Please use proper grouping symbols.
What you have is \(\displaystyle \frac{3x}{x}+2+\frac{2}{x}-1=\frac{5}{x^{2}}+x-2\)
I believe you mean:
\(\displaystyle \frac{3x}{x+2}+\frac{2}{x-1}=\frac{5}{x^{2}+x-2}\)
See?. Big difference without the proper grouping symbols. If the latter is correct, begin by factoring the quadratic and notice what it has in common with the other two denominators. Then, you'll have the LCD. As with any fraction, Make all denominators the same.
What you have is \(\displaystyle \frac{3x}{x}+2+\frac{2}{x}-1=\frac{5}{x^{2}}+x-2\)
I believe you mean:
\(\displaystyle \frac{3x}{x+2}+\frac{2}{x-1}=\frac{5}{x^{2}+x-2}\)
See?. Big difference without the proper grouping symbols. If the latter is correct, begin by factoring the quadratic and notice what it has in common with the other two denominators. Then, you'll have the LCD. As with any fraction, Make all denominators the same.
fasteddie65
Full Member
- Joined
- Nov 1, 2008
- Messages
- 360
banana said:I was wondering what the solutions to the rational equation 3x/x+2 + 2/x-1 = 5/x^2 + x - 2
Thank you
I assume you really mean: 3x/(x + 2) + 2/(x - 1) = 5/(x^2 + x - 2)
The solution I get graphically is x = -0.434....
Multiplying both sides by the LCD, I get 3x(x - 1) + 2(x + 2) = 5
3x^2 - 3x + 2x + 4 = 5
3x^2 - x - 1 = 0
x = 0.767... and x = -.0.434... are solutions from the Quadratic Formula.