How do you express #\frac { 5} { 3a } + 4b = \frac { 1- 2c} { d } # in terms of #c#?

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Answer 1 · 2017-08-09T14:49:12

#=>c=-(5d)/(6a)-(4bd)/2+1/2#

#5/(3a)+4b=(1-2c)/d#

#=>d*(5/(3a)+4b)=1-2c#

#=>(5d)/(3a)+4bd=1-2c#

#=> (5d)/(3a)+4bd-1=-2c#

#=> ((5d)/(3a)+4bd-1)/-2=c#

#=> (5d)/(3a*(-2))+(4bd)/-2-1/-2=c#

#=> (5d)/(-6a)-(4bd)/2+1/2=c#

#=>c=(5d)/(-6a)-(4bd)/2+1/2#

#=>c=-(5d)/(6a)-(4bd)/2+1/2#