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Answer :
The solution of the given rational equation [tex]\frac{x}{x+1}+\frac{3}{x+4}=\frac{x+3}{x+4}[/tex] is x = 0
In this question, we have been given a rational equation.
[tex]\frac{x}{x+1}+\frac{3}{x+4}=\frac{x+3}{x+4}[/tex]
We need to find the solution of the above rational equation.
[tex]\frac{x}{x+1}+\frac{3}{x+4}\\\\=\frac{x(x+4)+3(x+1)}{(x+1)(x+4)}\\\\ =\frac{x^{2}+4x+3x+3 }{(x+1)(x+4)} \\\\=\frac{x^{2} +7x+3}{(x+1)(x+4)}[/tex]
So given equation would be,
[tex]\frac{x^{2} +7x+3}{(x+1)(x+4)}=\frac{x+3}{x+4}\\\\x^{2} +7x+3=(x+1)(x+3)\\\\x^{2} +7x+3=x^{2} +x+3x+3\\\\x^{2} +7x+3=x^{2} +4x+3\\\\7x-4x=3-3\\\\3x=0\\\\x=0[/tex]
Therefore, the solution of the given rational equation [tex]\frac{x}{x+1}+\frac{3}{x+4}=\frac{x+3}{x+4}[/tex] is x = 0
Learn more about the rational equation here:
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