Question on: WAEC Mathematics - 2014
Subtract \(\frac{1}{2}\)(a - b - c) from the sum of \(\frac{1}{2}\)(a - b + c) and \(\frac{1}{2}\)
(a + b - c)
(a + b - c)
A
\(\frac{1}{2}\) (a + b + c)
B
\(\frac{1}{2}\) (a - b - c)
C
\(\frac{1}{2}\) (a - b + c)
D
\(\frac{1}{2}\) (a + b - c)
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Correct Option: A
\(\frac{1}{2}\)(a - b + c) + \(\frac{1}{2}\)(a + b - c) - [\(\frac{1}{2}\) (a - b - c)]
\(\frac{1}{2}a - \frac{1}{2}b + \frac{1}{2}c + \frac{1}{2}a + \frac{1}{2}b - \frac{1}{2}c - \frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)
= \(\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)
= \(\frac{1}{2}(a + b + c)\)
\(\frac{1}{2}a - \frac{1}{2}b + \frac{1}{2}c + \frac{1}{2}a + \frac{1}{2}b - \frac{1}{2}c - \frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)
= \(\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)
= \(\frac{1}{2}(a + b + c)\)
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