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# Solve for a in terms of b & c .

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Solve for a in terms of b & c.

(Your answer should say a = and the other side of the equation will have a b and c in it). (The last time I posted this I didn't have a specific math problem but now I do, sorry!)

Jan 31, 2019

#1
+3

so we can multiply both sides by abc

bc + ac = ab

ac - ab = -bc

a(c-b) = -bc

-a(c-b) = bc

-a = bc/(c-b)

so

a = bc/(c-b)

Jan 31, 2019
#2
+3

Solution:..

$$\large a=-\dfrac{bc}{c-b}\quad |\quad \:b\ne \:c$$

GA

Jan 31, 2019
#3
+4

$$\dfrac{1}{a} +\dfrac{1}{b}=\dfrac{1}c\\ \dfrac{1}a = \dfrac{1}c-\dfrac{1}b\\ \dfrac{1}a = \dfrac{b-c}{bc}\\ a = \dfrac{bc}{b-c}$$

.
Feb 1, 2019
#4
+6

Max, your way may be better. You need exceptions: $$b \ne c$$ & $$(b * c) \ne 0$$ .

...Solution:

$$\Large \frac{1}{a} \large abc+\Large \frac{1}{b} \large abc=\Large \frac{1}{c}\large abc\\ \large bc+ac=ab\\ \large bc+ac-bc=ab-bc\\ \large ac=ab-bc\\ \large ac-ab=ab-bc-ab\\ \large ac-ab=-bc\\ \large a\left(c-b\right)=-bc\\ \large \dfrac{a\left(c-b\right)}{c-b}=\dfrac{-bc}{c-b}\quad | \quad \:b\ne \:c\\ \text{ }\\ \LARGE a=-\dfrac{bc}{c-b}\quad | \quad \:b\ne \:c\\$$

GA

Feb 1, 2019
#5
0

a =bc / (b - c) is eqivalent to a =- bc /(c - b) !!!.........(1)
Let a=2, b =3, c =?

1/2 + 1/3 = 1/c
c = 6/5 . Sub these into (1) above:
2 =(3 * 6/5) /(3 - 6/5)
2 =3.6 / 1.8
2 = 2.
Now sub into the 2nd solution:
2=-(3 *6/5) / (6/5 - 3)
2 =-3.6 / - 1.8
2 = 2

Feb 1, 2019
#6
+5

Mr. BB, your equation would be correct if you leave off the triple-factorial (!!!).

a =bc / (b - c) is eqivalent to a =- bc /(c - b) !!!

I know you like these.  It does seem to be an ideal symbol for you –especially when used in this form: (!!!Mr. BB). Read as Triple Deranged Mr. BB

Until next time Mr. BB, keep up the Derangements!!!.

GA

GingerAle  Feb 1, 2019