3 years ago

Please help question #14!!!

1 answer:

7 0

$\text{The domain:\\a+3\neq0\ \wedge\ a-2\neq0\\a\neq-3\ \wedge\ a\neq2$

$\dfrac{5}{a+3}=\dfrac{3}{a-2}\ \ \ \ |\text{cross multiply}\\\\5(a-2)=3(a+3)\\\\5a-10=3a+9\ \ \ |+10\\\\5a=3a+19\ \ \ \ |-3a\\\\2a=19\ \ \ \ |:2\\\\a=9\dfrac{1}{2}\\\\Answer:\ B.\ a=9\dfrac{1}{2}$

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