The rational expression is given to be:
[tex]\frac{x-1}{x+1}-\frac{2x+3}{2x+1}[/tex]STEP 1: Find the Lowest Common Multiplier (LCM) of the denominators
[tex]\text{LCM of }(x+1)\text{ and }(2x+1)\Rightarrow(x+1)(2x+1)[/tex]STEP 2: Adjust the fractions by dividing the LCM by the denominator of each fraction and multiplying the numerator and denominator by the result
[tex]\begin{gathered} For\text{ }\frac{x-1}{x+1} \\ Multiplier=\frac{(x+1)(2x+1)}{x+1}=2x+1 \\ \text{New fraction:} \\ \Rightarrow\frac{(x-1)(2x+1)}{(x+1)(2x+1)} \end{gathered}[/tex]and
[tex]\begin{gathered} For\text{ }\frac{2x+3}{2x+1} \\ Multiplier=\frac{(x+1)(2x+1)}{2x+1}=x+1 \\ \text{New fraction:} \\ \Rightarrow\frac{(2x+3)(x+1)}{(2x+1)(x+1)} \end{gathered}[/tex]Hence, the expression becomes:
[tex]\Rightarrow\frac{(x-1)(2x+1)}{(x+1)(2x+1)}-\frac{(2x+3)(x+1)}{(2x+1)(x+1)}[/tex]STEP 3: Apply the fraction rule
[tex]\frac{a}{b}-\frac{c}{b}=\frac{a-c}{b}[/tex]Hence, the expression becomes:
[tex]\Rightarrow\frac{(x-1)(2x+1)-(2x+3)(x+1)}{(x+1)(2x+1)}[/tex]STEP 4: Expand and simplify the brackets using the FOIL method
[tex]\mleft(a+b\mright)\mleft(c+d\mright)=ac+ad+bc+bd[/tex]Hence, the expression becomes:
[tex]\begin{gathered} \Rightarrow\frac{(2x^2+x-2x-1)-(2x^2+2x+3x+3)}{2x^2+x+2x+1} \\ \Rightarrow\frac{(2x^2-x-1)-(2x^2+5x+3)}{2x^2+3x+1} \\ \Rightarrow\frac{2x^2-2x^2-x-5x-1-3}{2x^2+3x+1} \\ \Rightarrow\frac{-6x-4}{2x^2+3x+1} \end{gathered}[/tex]ANSWER:
The numerator is:
[tex]-6x-4[/tex]The denominator is:
[tex]2x^2+3x+1[/tex]