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rational equation and function
- Thread starter Rani
- Start date
Not quite. You made an error distributing the -3x to (x-1). In addition, always check your answers by plugging them back in to the original equation to see if they work and that there are no extraneous solutions.
\(\displaystyle \frac{-3x}{x+1}=\frac{-2}{x-1}\)
\(\displaystyle (-3x)(x-1)=(-2)(x+1)\)
\(\displaystyle -3x^2+3x=-2x-2\)
\(\displaystyle 3x^2-5x-2=0\)
\(\displaystyle (3x+1)(x-2)=0\)
\(\displaystyle 3x+1=0\) and \(\displaystyle x-2=0\)
therefore: \(\displaystyle x=\frac{-1}{3}\) and \(\displaystyle x=2\)
\(\displaystyle \frac{-3x}{x+1}=\frac{-2}{x-1}\)
\(\displaystyle (-3x)(x-1)=(-2)(x+1)\)
\(\displaystyle -3x^2+3x=-2x-2\)
\(\displaystyle 3x^2-5x-2=0\)
\(\displaystyle (3x+1)(x-2)=0\)
\(\displaystyle 3x+1=0\) and \(\displaystyle x-2=0\)
therefore: \(\displaystyle x=\frac{-1}{3}\) and \(\displaystyle x=2\)
Is this correct?
-3x/x+1=-2/x-1 \(\displaystyle \text{No. You must have grouping symbols around the denominators.}\)
\(\displaystyle -3x/(x + 1) = -2/(x - 1)\)
-3x(x-1)=-2(x+1)
-3x^2+3=-2x-2
-3x^2+2x+5=0
-(3x^2-2x-5)=0
-(3x-5)(x+1)=0
x=5/3
x=-1
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