waecmaths question:
Subtract $\tfrac{1}{2}(a-b-c)$ from the sum of $\tfrac{1}{2}(a-b+c)$and $\tfrac{1}{2}(a+b-c)$
Option A:
$\tfrac{1}{2}(a+b+c)$
Option B:
$\tfrac{1}{2}(a-b-c)$
Option C:
$\tfrac{1}{2}(a-b+c)$
Option D:
$\tfrac{1}{2}(a+b-c)$
waecmaths solution:
\[\begin{align} & [\tfrac{1}{2}(a-b+c)+\tfrac{1}{2}(a+b-c)]-[\tfrac{1}{2}(a-b-c)]=\tfrac{1}{2}[a-b+c+a+b-c]-[\tfrac{1}{2}(a-b-c)] \\ & =\tfrac{1}{2}[2a]-\tfrac{1}{2}[a-b-c] \\ & =\tfrac{1}{2}[2a-a+b+c] \\ & =\tfrac{1}{2}[a+b+c] \end{align}\]
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