Given that a+b=7 and a-b=3 find: 3^a/3^b

1 answer:

8 0

$\bf \begin{cases} a+b=7\\ \boxed{b}=7-a\\[-0.5em] \hrulefill\\ a-b=3 \end{cases}\implies \stackrel{\textit{substituting \underline{b} in the 2nd equation}}{a-\left( \boxed{7-a} \right)=3}\implies 2a-7=3 \\\\\\ 2a=10\implies a=\cfrac{10}{2}\implies \blacktriangleright a=5 \blacktriangleleft \\\\\\ \stackrel{\textit{substituting \underline{a} in the 1st equation}}{5+b=7\implies \blacktriangleright b=2 \blacktriangleleft} \\\\[-0.35em] ~\dotfill$

$\bf \cfrac{3^a}{3^b}\implies \cfrac{3^5}{3^2}\implies 3^5\cdot 3^{-2}\implies 3^{5-2}\implies 3^3\implies 27$

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(A)

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