Giving brainliest to anyone who answers
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Answer:
[tex]\frac{1}{x+2}-\frac{1}{3} =\frac{1}{x^2 +5x+6}[/tex]
Step-by-step explanation:
\frac{1}{x+2}-\frac{1}{3} =\frac{1}{x^2 +5x+6}
Answer:
[tex]$\frac{2}{3x} +\frac{7}{5y} $[/tex]
[tex]$\frac{2(5y)}{3x(5y)} +\frac{7(3x)}{5y(3x)} $[/tex]
[tex]$\frac{10y}{15xy} +\frac{21x}{15xy}$ \\[/tex]
[tex]$\frac{10y+21x}{15xy}$ \\[/tex]
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[tex]$\frac{1}{x+2} -\frac{1}{x+3} $[/tex]
[tex]$\frac{1(x+3)}{(x+2)(x+3)} -\frac{1(x+2)}{(x+2)(x+3)} $[/tex]
[tex]$ \frac{x+3- (x+2)}{(x+2)(x+3)} $[/tex]
[tex]$ \frac{1}{(x+2)(x+3)} $[/tex]