$$\frac{1+3}{7}-\frac{2r-5}{3}=\frac{3n-5}{6}$$
$n=-\frac{4r}{3}+\frac{43}{7}$
$$6\left(1+3\right)-14\left(2r-5\right)=7\left(3n-5\right)$$
$$6\times 4-14\left(2r-5\right)=7\left(3n-5\right)$$
$$24-14\left(2r-5\right)=7\left(3n-5\right)$$
$$24-28r+70=7\left(3n-5\right)$$
$$94-28r=7\left(3n-5\right)$$
$$94-28r=21n-35$$
$$21n-35=94-28r$$
$$21n=94-28r+35$$
$$21n=129-28r$$
$$\frac{21n}{21}=\frac{129-28r}{21}$$
$$n=\frac{129-28r}{21}$$
$$n=-\frac{4r}{3}+\frac{43}{7}$$
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$r=-\frac{3n}{4}+\frac{129}{28}$
$$-28r=21n-35-94$$
$$-28r=21n-129$$
$$\frac{-28r}{-28}=\frac{21n-129}{-28}$$
$$r=\frac{21n-129}{-28}$$
$$r=-\frac{3n}{4}+\frac{129}{28}$$
