Find b-a if the graph of $Ax + By = 3$ passes through $(-7,2)$ and is parallel to the graph of $x + 3y = -5.$
From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B. Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.
Then the line is y = -x/3 - 19/3. Then 3y = -x - 19, so 3y + x = -19. We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3. Therefore, A + B = -9/19 - 3/19 = -12/19.
Find a linear inequality with the following solution set. Each grid line represents one unit.
https://latex.artofproblemsolving.com/4/8/2/48234b947b623e95b3f920ee2e9ce6cb24dcfa4b.png
Give your answer in the form $ax+by+c>0$ or $ax+by+c\geq0$ where a and b and c are integers with no common factor greater than 1.)
We have the points (-1,-1) and (1,0) on this graph if we had a solid line
The slope of this line is [ 0 - - 1 ] / [ 1 - -1] = [1]/ [2] = 1/2
So....using the point (1,0) and the slope we have either that
y < (1/2) ( x - 1) or y > (1/2)( x - 1)
Since the point (0,0) is in the feasible region, note that 0 > (1/2) ( 0 - 1) = 0 > -1/2 is true
So....the inequality is
y > (1/2) ( x - 1)
y > (1/2)x - 1/2
(1/2)x - y - 1/2 < 0 multiply through by 2
x - 2y - 1 < 0