Can anyone teach me how do you get from right side to left side equation?
- Thread starter bbqqcc123
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What did you get, when you attempted to solve for [imath]y=[/imath] by the usual methods?
a) You converted the constant [imath]e^{4C}[/imath] to a simpler notation; that is, to [imath]\tilde{C}[/imath].
b) You multiplied through to get rid of the denominator.
c) You subtracted to get the [imath]y[/imath]-containing terms on one side of the "equals" sign.
d) You factored out the [imath]y[/imath].
e) ....
Please reply showing all of your steps. Thank you!
BigBeachBanana
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[math]\dfrac{y-2}{y+2} = \dfrac{y+2-2-2}{y+2}=1- \dfrac{4}{y+2}[/math]
Looks like I belong to the ones who don't XD. Thank you so much for the help, appreciate it!
Assume:
\(\displaystyle \dfrac {a}{b} \ = \ \dfrac {c}{d} \) →
\(\displaystyle \dfrac {a}{b} + 1 \ = \ \dfrac {c}{d} + 1 \) →
\(\displaystyle \dfrac {a+b}{b} \ = \ \dfrac {c+d}{d} \) ......................................(1) →
Then
\(\displaystyle \dfrac {a}{b} \ = \ \dfrac {c}{d} \) →
\(\displaystyle \dfrac {a}{b} - 1 \ = \ \dfrac {c}{d} - 1 \) →
\(\displaystyle \dfrac {a - b}{b} \ = \ \dfrac {c - d}{d} \) ......................................(2)
Divide (1) by (2) to get
\(\displaystyle \dfrac {a+b}{a - b} \ = \ \dfrac {c + d}{c - d} \) ........................(3)
apply (3) into the given equation to get the desired result
\(\displaystyle \dfrac {a}{b} \ = \ \dfrac {c}{d} \) →
\(\displaystyle \dfrac {a}{b} + 1 \ = \ \dfrac {c}{d} + 1 \) →
\(\displaystyle \dfrac {a+b}{b} \ = \ \dfrac {c+d}{d} \) ......................................(1) →
Then
\(\displaystyle \dfrac {a}{b} \ = \ \dfrac {c}{d} \) →
\(\displaystyle \dfrac {a}{b} - 1 \ = \ \dfrac {c}{d} - 1 \) →
\(\displaystyle \dfrac {a - b}{b} \ = \ \dfrac {c - d}{d} \) ......................................(2)
Divide (1) by (2) to get
\(\displaystyle \dfrac {a+b}{a - b} \ = \ \dfrac {c + d}{c - d} \) ........................(3)
apply (3) into the given equation to get the desired result
thx!