Answer :

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]

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Rewritten by : Barada