Answer:
We conclude that:
[tex]c=-\frac{5n}{3\left(-n+2\right)};\quad \:n\ne \:2[/tex]
Step-by-step explanation:
Given the expression
[tex]\frac{4}{2n}+\frac{5}{3c}=1[/tex]
Let us solve for 'c'
[tex]\frac{4}{2n}+\frac{5}{3c}=1[/tex]
Least Common Multiplier of 2n, 3c: 6nc
Now multiply both sides by LCM = 6nc
[tex]\frac{4}{2n}\cdot \:6nc+\frac{5}{3c}\cdot \:6nc=1\cdot \:6nc[/tex]
Simplify
[tex]12c+10n=6nc[/tex]
Subtract 10n from both sides
[tex]12c+10n-10n=6nc-10n[/tex]
Simplify
[tex]12c=6nc-10n[/tex]
Subtract 6nc from both sides
[tex]12c-6nc=6nc-10n-6nc[/tex]
Simplify
[tex]12c-6nc=-10n[/tex]
Factor 12c - 6nc = 6c(2-n)
[tex]6c\left(2-n\right)=-10n[/tex]
Divide both sides by 6(2-n); n≠2
[tex]\frac{6c\left(2-n\right)}{6\left(2-n\right)}=\frac{-10n}{6\left(2-n\right)};\quad \:n\ne \:2[/tex]
simplify
[tex]c=-\frac{5n}{3\left(-n+2\right)};\quad \:n\ne \:2[/tex]
Therefore, we conclude that:
[tex]c=-\frac{5n}{3\left(-n+2\right)};\quad \:n\ne \:2[/tex]
