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Find b-a if the graph of $Ax + By = 3$ passes through $(-7,2)$ and is parallel to the graph of $x + 3y = -5.$

 Apr 7, 2020
 #1
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+1

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.

 Apr 7, 2020
 #2
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0

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.)

 Apr 7, 2020
 #3
avatar+111329 
+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

 

 

cool cool cool

CPhill  Apr 7, 2020

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