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avatar+73 

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
avatar+148 
+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
avatar+2234 
+1

Solution:..

Answer:  

 

a=bccb|bc

 

 

 

 

GA

 Jan 31, 2019
 #3
avatar+9675 
+4

1a+1b=1c1a=1c1b1a=bcbca=bcbc

.
 Feb 1, 2019
 #4
avatar+2234 
+4

Max, your way may be better. You need exceptions: bc & (bc)0 .   

 

...Solution: 

 

1aabc+1babc=1cabcbc+ac=abbc+acbc=abbcac=abbcacab=abbcabacab=bca(cb)=bca(cb)cb=bccb|bc a=bccb|bc

 

 

GA

 Feb 1, 2019
 #5
avatar
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
avatar+2234 
+3

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

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