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Complex Fractions
- Thread starter ktrpies
- Start date
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1/(m-1) + 2/(m+2)
----------------------
2/(m+2) - 1/(m-3)
doing the top line first
1(m+2)/ (m-1)(m+2) + 2(m-1) / (m-1) (m+2)
=m+2 +2(m-1) / (m+2) (m-1)
= 3m / (m+2) (m-1)
now the bottom line
2/(m+2) - 1/(m-3)
=2 (m-3) / (m+2) (m-3) - 1 ( m+2) / (m+2) (m-3)
= 2(m-3) - 1(m+2) / (m+2) (m-3)
= 2m - 6 -m -2 / (m+2) (m-3)
= 3m -8 / (m+2) (m-3)
bring the two parts back together and cancel out the common parts and you have the answer.
----------------------
2/(m+2) - 1/(m-3)
doing the top line first
1(m+2)/ (m-1)(m+2) + 2(m-1) / (m-1) (m+2)
=m+2 +2(m-1) / (m+2) (m-1)
= 3m / (m+2) (m-1)
now the bottom line
2/(m+2) - 1/(m-3)
=2 (m-3) / (m+2) (m-3) - 1 ( m+2) / (m+2) (m-3)
= 2(m-3) - 1(m+2) / (m+2) (m-3)
= 2m - 6 -m -2 / (m+2) (m-3)
= 3m -8 / (m+2) (m-3)
bring the two parts back together and cancel out the common parts and you have the answer.
Hello, ktrpies!
. . . .1 . . . . . 2
. . ------- + ------- . (m - 1)(m + 2)(m - 3)
. . m - 1 . . m + 2 . . . . . . . . . . . . . . . . . . . . . . .(m + 2)(m - 3) + 2(m - 1)(m - 3)
. . ------------------ . . . . . . . . . . . . . . . . . . . = . -----------------------------------------
. . . . 2 . . . . .1 . . . . . . . . . . . . . . . . . . . . . . . . . .2(m - 1)(m - 3) - (m - 1)(m + 2)
. . -------- - ------- . (m - 1)(m + 2)(m - 3)
. . m + 2 . . m - 3
. . . . . .m<sup>2</sup> - m - 6 + 2m<sup>2</sup> - 8m + 6 . . . . . . 3m<sup>2</sup> - 9m . . . . . . . 3m(m - 3)
. . = . ----------------------------------- . = . ----------------- . = . -----------------
. . . . . .2m<sup>2</sup> - 8m + 6 - m<sup>2</sup> - m + 2 . . . . . m<sup>2</sup> - 9m + 8 . . . . .(m - 1)(m - 8)
Multiply top and bottom by the LCD: (m - 1)(m + 2)(m - 3)Simplify:
1/(m-1) + 2/(m+2)
2/(m+2) - 1/(m-3)
Anwer:
3m(m-3)
(m-1)(m-8)
. . . .1 . . . . . 2
. . ------- + ------- . (m - 1)(m + 2)(m - 3)
. . m - 1 . . m + 2 . . . . . . . . . . . . . . . . . . . . . . .(m + 2)(m - 3) + 2(m - 1)(m - 3)
. . ------------------ . . . . . . . . . . . . . . . . . . . = . -----------------------------------------
. . . . 2 . . . . .1 . . . . . . . . . . . . . . . . . . . . . . . . . .2(m - 1)(m - 3) - (m - 1)(m + 2)
. . -------- - ------- . (m - 1)(m + 2)(m - 3)
. . m + 2 . . m - 3
. . . . . .m<sup>2</sup> - m - 6 + 2m<sup>2</sup> - 8m + 6 . . . . . . 3m<sup>2</sup> - 9m . . . . . . . 3m(m - 3)
. . = . ----------------------------------- . = . ----------------- . = . -----------------
. . . . . .2m<sup>2</sup> - 8m + 6 - m<sup>2</sup> - m + 2 . . . . . m<sup>2</sup> - 9m + 8 . . . . .(m - 1)(m - 8)
apm said:now the bottom line
2/(m+2) - 1/(m-3)
=2 (m-3) / (m+2) (m-3) - 1 ( m+2) / (m+2) (m-3)
= 2(m-3) - 1(m+2) / (m+2) (m-3)
= 2m - 6 -m -2 / (m+2) (m-3)
= 3m -8 / (m+2) (m-3) : typo; numerator should be m - 8
1/(m-1) + 2/(m+2)
----------------------
2/(m+2) - 1/(m-3)
Just another way; I use it cause I hate typing!
a = m-1
b = m+2
c = m-3
So equation can be rewritten this way:
1/a + 2/b
----------- =
2/b - 1/c
(b + 2a) / ab
--------------- =
(2c - b) / bc
bc(b + 2a)
------------ =
ab(2c - b)
c(b + 2a)
----------- =
a(2c - b)
2ac + bc
----------
2ac - ab
Now substitute back and you'll get soroban's answer.
----------------------
2/(m+2) - 1/(m-3)
Just another way; I use it cause I hate typing!
a = m-1
b = m+2
c = m-3
So equation can be rewritten this way:
1/a + 2/b
----------- =
2/b - 1/c
(b + 2a) / ab
--------------- =
(2c - b) / bc
bc(b + 2a)
------------ =
ab(2c - b)
c(b + 2a)
----------- =
a(2c - b)
2ac + bc
----------
2ac - ab
Now substitute back and you'll get soroban's answer.
Thank you guys. Very helpful! On my first try I did the numerator and the denominator seperately and found the LCD's for each. Then I got hasty and started cancelling terms, not factors, which lead me astray. The second time I found the LCD for the entire operation, but did not factor or group correctly. Finally, I got the technique right but had one mistake at the begining which threw my answer off. But at least I know my mistakes now, thanks to your guidence. Thanks for the speed of the replies also!