Respuesta :
well, keeping in mind that parallel lines have the same exact slope, hmmmmm
[tex]\bf y = \stackrel{\stackrel{m}{\downarrow }}{2}x+1\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
alrite, so we know the slope has to be 2 then hmmm let's check the point it runs through
[tex]\bf \begin{cases} 3x-2y=&10\\\\ x+y=&5\\ \qquad y=&5-x \end{cases} \\\\\\ \stackrel{\textit{substituting \underline{y} on the 1st equation}}{3x-2(5-x)=10}\implies 3x-10+2x=10\implies 5x-10=10 \\\\\\ 5x=20\implies x=\cfrac{20}{5}\implies \blacktriangleright x = 4 \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{y = 5-x}\implies y=5-4\implies \blacktriangleright y = 1 \blacktriangleleft[/tex]
so, we're really looking for the equation of a line whose slope is 2 and runs through (4,1)
[tex]\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{1})~\hspace{10em} \stackrel{slope}{m}\implies 2 \\\\\\ \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}{2}(x-\stackrel{x_1}{4}) \\\\\\ y-1=2x-8\implies y=2x-7[/tex]
