Answer:
[tex]\displaystyle A=\frac{x-2}{x+3}[/tex]
[tex]\displaystyle B=\frac{-5}{(x+3)(x+2)}[/tex]
Step-by-step explanation:
We need to find two expressions with unlike denominators what sum
[tex]\displaystyle S=\frac{x-3}{x+2}[/tex]
Let's suppose one of the expressions is:
[tex]\displaystyle A=\frac{x-2}{x+3}[/tex]
Now we subtract S minus A to find the other expression B:
[tex]\displaystyle B=S-A=\frac{x-3}{x+2}-\frac{x-2}{x+3}[/tex]
Multiply the first fraction by x+3 and the second by x+2;
[tex]\displaystyle B=(x+3)\frac{x-3}{(x+3)(x+2)}-(x+2)\frac{x-2}{(x+3)(x+2)}[/tex]
Operating:
[tex]\displaystyle B=\frac{x^2-9}{(x+3)(x+2)}-\frac{x^2-4}{(x+3)(x+2)}[/tex]
Subtracting both fractions with like denominators:
[tex]\displaystyle B=\frac{x^2-9-(x^2-4)}{(x+3)(x+2)}[/tex]
Simplifying:
[tex]\displaystyle B=\frac{-5}{(x+3)(x+2)}[/tex]
Thus the two expressions are:
[tex]\displaystyle A=\frac{x-2}{x+3}[/tex]
And
[tex]\displaystyle B=\frac{-5}{(x+3)(x+2)}[/tex]
