How do you solve #2( 4r + 6) = \frac { 12} { 5} n - 2n + 3#?

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Answer 1 · 2017-11-03T04:56:23

#n= (40r+45)/2#
or
#r = (2n-45)/40#

#2( 4r + 6) = \frac { 12} { 5} n - 2n + 3#

Apply BODMAS rule and transposition method :

# 8r +12 = \frac { 12} { 5} n - 2n + 3#

# 8r +12-3=\frac {12} {5} n-2nxx(5/5)#----for equating denominators

#8r +9 = (12n-10n)/5#

#8r+9 = (2n)/5#

# 5xx(8r+9) = 2n#

# 40 r + 45 = 2n#

#therefore n= (40r+45)/2#

or

# 40 r + 45 = 2n#

#=> 40r = 2n-45#

#therefore r = (2n-45)/40#