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Here, the least common multiple of 4cx2, 2x, and Sacs ; (Art. 147), is Saco x2 ; then, Sac x 2
X Sa'b=2ac + 3a2b=Ca'bc 4cxa Bacara
new numeraXy=4aco Xy=4acy 2x Sac? x2
X 5x2=X 5x2=524 8ac2
are the 8ac2x2' Sacox?? 8acx2 fractions required.
1 Ex. 6. Reduce
and to a comat 3
2.0 mon denominator. 30x2 2a2x - 2x3
3a + 3x Ans.
and Gax+6x2 Gar+6x? Gax +6x2
2x+3 5.0 +2 Ex. 7. Reduce
to a com
3ab mon denominator.
6abx +9ab 5x2 + 2x Ans.
Ex. 8. Reduce
3' mon denominator.
4x2 +42 Ans.
За — х2
b' d' common denominator.
adx 2bc x Ans.
and bdx' bdx
40-15 Ex. 10. Reduce
and 7+ 54' 3x
2 to a common denominator.
6x9 30axy-10~2 y + 50y Ans.
5& Ex, 11. Reduce
and 4a2-4x3' 4a2 - 4x2?
1 Ex. 12. Reduce
and a2 + 2ax + x2 5y
to the least common denominator. -24
a3 +ara + ax + x3 as-ax4 ta' x a' - ax4 t-a4 x 75
-a x4 tra4x
§ V. ADDITION AND SUBTRACTION OF ALGEBRAIC
To add fractional quantities together.
154. Reduce the fractions, if necessary, to a common denominator, by the rules in the last case, then add all the numerators together, and under their sum put the common denominator; bring the resulting fraction to its lowest terms, and it will be the sum required.
20 5x Ex. 1. Add
3' 7 ? 9 2x X7X9=126x
126x +135x + 21x_282x 5x X 3X9=13500
139 X7 X3= 21x
is the sum required.
189 3X7X9= 189
56 Ex. 2. Add
and together. Õ 36 4a
12ao b+8ab-+-1563 a X 36 X 4a=12026
12ab2 2a X6 X 4a= 8a2b
-(dividing by 6) 56 x 36 xb= 1563 12ab2
is the sum b X 36 X 4a=12ab2 12ab
Or, the least common multiple of the denominators may be found, and then proceed, as in (Art.
It is generally understood that mixed quantities are reduced to improper, fractions, before we perform any of the operations of Addition and Subtraction. But it is best to bring the fractional parts only to a common denominator, and to affix their sum or difference to the sum or difference of the integral part interposing the proper sign.
30* Ex. 3. It is required to find the sum of a
b 2ar and bt
2ax bc + 2ax Here, a
and 6+ 6
6 Then, (ab — 3x2) xc=abc - 3cx 2
numerators. (bc+ 2ax) xbbc-2abx
& bc 2abx - 3cx?
2abx - 3era zatot bc
bc is the sum required.
Or, bringing the fractional parts only to a common denominator, Thus, 302 Xc=3cx2
numerators, 2ax Xb=2abx And bxc=bc common denominator. 3cx2 2abx
2abx - 3cx Whence a
+6+ =a+b+1 bc
be the sum.
Ex. 4. It is required to find the sum of 5:0+ -2
2x - 3 and 4x 3
5x Here, ( *—2) X 5x=5x2 -- 10x
numerators. (x-3) X3 =6x - 9
And 3 X 5x=15x common denominator. 5x2-10x
6x -9 Whence 5x --
=9x + 150
15x 5x +-10%, 9-6.0 5x2-16x+9 + =9x+ to
the sum 150 153
15x required. 6x -9
9-62 Here, is evidently
(Art. 128); 153
15.30 but we might change the fractions into other equivalent forms before we begin to add or subtract ; thus, the fractional part of the proposed quantity
2x - 3 4x
may be transformed by changing the
5x signs of the numerator, (Art. 128), and the quantity
3-260 itself can be written thus, 4x+ : It is well to
5x keep this transformation in mind, as it is often necessary to make use of it in performing several algebraical operations.
3a2 2a b Ex. 5. Add and together. 26' 5
105a +-28ab +1002 Ans.