Solve simultaneously..? #x = 3^y# and #x = 1/2 (3 + 9y)#

1 Answer
Nov 23, 2017

Answer:

This is the method I used in deducing the following simultaneously equation..

See steps below;

Explanation:

Solving simultaneously..

#x = 3^y - - - - - - eqn1#

#x = 1/2(3 + 9y) - - - - - - eqn2#

Look at the common value in both equations..

#x# is the common, hence we equate both together..

Having..

#3^y = 1/2(3 + 9y)#

#3^y = (3 + 9y)/2#

Cross multiplying..

#3^y/1 = (3 + 9y)/2#

#2xx 3^y = 3 + 9y#

#6^y = 3 + 9y#

Log both sides..

#log6^y = log(3 + 9y)#

Recall the law of logarithm #-> log6^y = x, ylog6 = x#

Therefore...

#ylog6 = log(3 + 9y)#

Divide both sides by #log6#

#(ylog6)/(log6) = log(3 + 9y)/(log6)#

#(ycancel(log6))/cancel(log6) = log(3 + 9y)/(log6)#

#y = (log(3 + 9y))/log(6)#

#y = (cancel (log)(3 + 9y))/(cancel (log)(6))#

#y = (3 + 9y)/6#

Cross multiplying..

#y/1 = (3 + 9y)/6#

#6 xx y = 3 + 9y#

#6y = 3 + 9y#

Collect like terms

#6y - 9y = 3#

#-3y = 3#

Divide both sides by #-3#

#(-3y)/(-3) = 3/-3#

#(cancel(-3)y)/cancel(-3) = 3/-3#

#y = -3/3#

#y = - 1#

Substitute the value of #y# into #eqn1# to get #x#

#x = 3^y - - - - - - eqn1#

#x = 3^-1#

Recall in indices, #x^-1 = 1/x#

#:. x = 1/3#

Hence the values are #rArr x = 1/3, y = -1#

Hope this helps!