更新1:
its solving rational equations
how do you solve this problem?
回答 (7)
2009-06-08 5:16 am
✔ 最佳答案
OK 5/c = 2/ (c+3) ; cross multiply
5(c+3) = 2c
5c+15 = 2c
15 = -3c
-5 = c
Proof
5/-5 = 2(-5+3) ??
-1 = 2/-2 ??
-1 = -1 YES
Hope that helps.
2009-06-08 5:17 am
With any rational equation (equation with fractions), I would always cross multiply first.
5(c+3) = 2c
5c+15=2c
3c=-15
c=-5.
Life is easier without fractions.
5(c+3) = 2c
5c+15=2c
3c=-15
c=-5.
Life is easier without fractions.
參考: No source.
2009-06-08 5:14 am
First get the unknown onto one side:
5/c = 2/(c+3)
(c+3)/c = 2/5
Then simplify:
c/c + 3/c = 2/5
1 + 3/c = 2/5
3/c = -3/5
1/c = -1/5
c = -5
5/c = 2/(c+3)
(c+3)/c = 2/5
Then simplify:
c/c + 3/c = 2/5
1 + 3/c = 2/5
3/c = -3/5
1/c = -1/5
c = -5
2009-06-08 5:31 am
5/c = 2/(c + 3)
5 = c[2/(c + 3)]
5 = 2c/(c + 3)
5(c + 3) = 2c
5*c + 5*3 = 2c
5c + 15 = 2c
5c - 2c = -15
3c = -15
c = -15/3
c = -5
5 = c[2/(c + 3)]
5 = 2c/(c + 3)
5(c + 3) = 2c
5*c + 5*3 = 2c
5c + 15 = 2c
5c - 2c = -15
3c = -15
c = -15/3
c = -5
2009-06-08 5:26 am
Cross-multiply:
Multiply the numerator of the left side (5) by the denominator of the right side (c+3) = (5)*(c+3) = 5c+15.
Multiply the denominator of the left side (c) by the numerator of the right side (2) = 2c.
Then, 5c+15 = 2c, 15 = 2c-5c, 15 = -3c, c = -5.
Multiply the numerator of the left side (5) by the denominator of the right side (c+3) = (5)*(c+3) = 5c+15.
Multiply the denominator of the left side (c) by the numerator of the right side (2) = 2c.
Then, 5c+15 = 2c, 15 = 2c-5c, 15 = -3c, c = -5.
2009-06-08 5:20 am
2c=5c+15
-3c=15
c=-5
-3c=15
c=-5
2009-06-08 5:16 am
5/c = 2/(c+3)
5(c+3)/c = 2
5(c+3) = 2c
5c + 15 = 2c
5c - 2c +15 = 0
3c = -15
c = -5
5(c+3)/c = 2
5(c+3) = 2c
5c + 15 = 2c
5c - 2c +15 = 0
3c = -15
c = -5
收錄日期: 2021-05-01 12:28:37
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