\fdO'^ - Old Forge Coal Mines

ARITHMETIC. 21
fractions to fractions having a common denominator. Then,
subtract the numerators, and place the remainder over the
common denominator.
7X2 14 14 7 14-7 7
Ans.
it) lU 10 10 lU
» X '/«
7
{U) 13 — 7— = ? This problem may
ways:
be solved in two
First: 13 = 12^, since ^ = 1, and 12^1 = 12 + ^ =
32 + 1 = 13.
12|f We can now subtract the whole numbers sepa-
7y\ rately, and the fractions separately, and obtain 12 — 7
TV K ^16 7 lG-7 9 ^,9 ^9 .
5t^ =5and--- = -3^ = -. 5 + - = 5-. Ans.
Second : By reducing both numbers to improper fractions
having a denominator of 16.
13
_2_3 X 16 208 7 ^ ^ (7 X 16) + 7 _ 112 + 7 ~ ~ ~
~" _
1 1 X 16 16 16 16
16
~
119
16'
o u . .• 1. 208 119 208 - 119 89 ,
Subtractmg, ^' we have —
16
—- =
16
—
16
= --
16
and
89 - -...Nr,/^/^ o = 16)89(5^
oO
the same result
fi,3,^^,hod.
that was obtained by the
9 (