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

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

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

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

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

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

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

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

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

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

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

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