What is the equation of this line in standard form?6x−5y=−136x−7y=116x−7y=−117x−6y=11

keeping in mind that standard form for a linear equation means• all coefficients must be integers, no fractions• only the constant on the right-hand-side• all variables on the left-hand-side, sorted• "x" must not have a negative coefficient[tex]\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{\frac{1}{2}}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{\frac{1}{2}}-\underset{x_1}{(-3)}}}\implies \cfrac{2+1}{\frac{1}{2}+3}\implies \cfrac{3}{\frac{1+6}{2}}\implies \cfrac{\frac{3}{1}}{~~\frac{7}{2}~~}\implies \cfrac{3}{1}\cdot \cfrac{2}{7}\implies \cfrac{6}{7}[/tex][tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{6}{7}}[x-\stackrel{x_1}{(-3)}]\implies y+1=\cfrac{6}{7}(x+3) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{7}}{7(y+1)=7\left( \cfrac{6}{7}(x+3) \right)}\implies 7y+7=6(x+3)\implies 7y+7=6x+18 \\\\\\ 7y=6x+11\implies -6x+7y=11\implies 6x-7y=-11[/tex]