Find the value of if the graph of passes through the point and is parallel to the graph of
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\(x+3y=-5 \)
\(3y=-x-5\)
\(y=-\frac{1}{3}x-\frac{5}{3}\)
so the slope of the line that is parallel to \(x+3y=-5\) is \(-\frac{1}{3}\)
Using point-slope form, we have:
\(y+7=-\frac{1}{3}(x-2)\)
\(y=-\frac{1}{3}x-\frac{19}{3}\)
\(3x+3y=-19\)
then, you multiply A and B by -3/19. obviously, B-A will equal 0.
JP