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

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Answer 1 · 2017-08-09T14:49:12

$\Rightarrow c = - \frac{5 d}{6 a} - \frac{4 b d}{2} + \frac{1}{2}$ $=>c=-(5d)/(6a)-(4bd)/2+1/2$

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

$\Rightarrow d \cdot (\frac{5}{3 a} + 4 b) = 1 - 2 c$ $=>d*(5/(3a)+4b)=1-2c$

$\Rightarrow \frac{5 d}{3 a} + 4 b d = 1 - 2 c$ $=>(5d)/(3a)+4bd=1-2c$

$\Rightarrow \frac{5 d}{3 a} + 4 b d - 1 = - 2 c$ $=> (5d)/(3a)+4bd-1=-2c$

$\Rightarrow \frac{\frac{5 d}{3 a} + 4 b d - 1}{- 2} = c$ $=> ((5d)/(3a)+4bd-1)/-2=c$

$\Rightarrow \frac{5 d}{3 a \cdot (- 2)} + \frac{4 b d}{- 2} - \frac{1}{- 2} = c$ $=> (5d)/(3a*(-2))+(4bd)/-2-1/-2=c$

$\Rightarrow \frac{5 d}{- 6 a} - \frac{4 b d}{2} + \frac{1}{2} = c$ $=> (5d)/(-6a)-(4bd)/2+1/2=c$

$\Rightarrow c = \frac{5 d}{- 6 a} - \frac{4 b d}{2} + \frac{1}{2}$ $=>c=(5d)/(-6a)-(4bd)/2+1/2$

$\Rightarrow c = - \frac{5 d}{6 a} - \frac{4 b d}{2} + \frac{1}{2}$ $=>c=-(5d)/(6a)-(4bd)/2+1/2$

How do you express \frac { 5} { 3a } + 4b = \frac { 1- 2c} { d } 53a+4b=1−2cd\frac { 5} { 3a } + 4b = \frac { 1- 2c} { d } in terms of ccc?