Express as a single fraction:

`(x-1)/(2x^2 -x-3)-(x+2)/(2x^2 +x-6)`

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`(x-1)/(2x^2-x-3)-(x+2)/(2x^2+x-6)`

To express this as single fraction, factor the denominators.

`=(x-1)/((x+1)(2x-3)) -(x+2)/((x+2)(2x-3))`

Notice that in the second fraction x+2 is present both on the numerator and denominator. So, cancel its common factor to simplify.

`=(x-1)/((x+1)(2x-3))-1/(2x-3)`

Then, multiply all the different factors in the denominators to get the LCD.

`(x+1)*(2x-3)=(x+1)(2x-3)`

Hence, the LCD of the two fractions is (x+1)(2x-3).

Since the denominator of the first fraction contains all the factors of the LCD, then it is only the equivalent fraction of 1/(2x-3) must be determined.

To get its equivalent fraction, notice its denominator have only one of the factor of the LCD which is (2x-3). So to have a denominator od (x+1)(2x-3), multiply its numerator and denominator by the missing factor (x+1).

`=(x-1)/((x+1)(2x-3))-1/(2x-3)*(x+1)/(x+1)`

`=(x-1)/((x+1)(2x-3))-(x+1)/((x+1)(2x-3))`

Now that they have same denominators, subtract them.

`= ((x-1)-(x+1))/((x+1)(2x-3))`

`=(x-1-x-1)/((x+1)(2x-3))`

`=(-2)/((x+1)(2x-3))`

`=-2/((x+1)(2x-3))`

**Hence, `(x-1)/(2x^2-x-3)-(x+2)/(2x^2+x-6)=-2/((x+1)(2x-3))` .**

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