Answer:
Computing the sum of fractions on the left side of the equation, we get:
[tex]\frac{x(x+5)+3(x+4)}{(x+4)(x+5)}=\frac{x+2}{x^2+9x+20}.[/tex]
Now, notice that:
[tex]x^2+9x+20=(x+4)(x+5).[/tex]
Therefore, we can rewrite the equation as:
[tex]\frac{x(x+5)+3(x+4)}{(x+4)(x+5)}=\frac{x+2}{(x+4)(x+5)}.[/tex]
The above result implies that:
[tex]x(x+5)+3(x+4)=x+2.[/tex]
Simplifying the above equation, we get:
[tex]\begin{gathered} x^2+5x+3x+12=x+2, \\ x^2+8x+12=x+2. \end{gathered}[/tex]
Subtracting x+2, we get:
[tex]x^2+7x+10=0.[/tex]
Notice that:
[tex]x^2+7x+10=(x+2)(x+5).[/tex]
Therefore,
[tex]x=-2,\text{ }[/tex]
because x≠-5, otherwise we could not have a denominator of x+5.
Answer:
[tex]x=2.[/tex]
