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Given that \(-3 \le 7x + 2y \le 3\) and \(-4 \le y - x \le 4\) , what is the largest possible value of x + y?

 

Can someone show me a step by step solution? I was thinking that I can solve this like a normal equation, and subsitute y. Am I right? Thanks.

 Nov 18, 2019
 #1
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Please do not discuss this problem!  This is an active homework problem.

 

To the original poster: I realize that homework may be challenging. If you wish to receive some help from the staff or other students, I encourage you to use the resources that the online classes provide, such as the Message Board.  Thanks.

 Nov 19, 2019
 #2
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Who are you to tell me not to help this kid? 

The only reason I don't is that I don't know how. 

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Guest Nov 19, 2019
 #3
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I agree, Wonderman, you do not have the authority to do this.

 

The question asker may have not recieved sufficient help.

CalculatorUser  Nov 19, 2019
 #4
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Notice if you isolate y in both equations, you can get four different linear inequalities that you can graph.

 

So isolating:

\(\frac{-3-7x}{2}\leq{y}{\leq}\frac{3-7x}{2}\)

 

\(-4+x\leq{y}{\leq}4+x\)

 

Ok now we have:

 

 

Graph:

 

Notice the intersection shape of the FOUR inequality inequations.

 

The MAXIMUM value of x + y, which is the answer you are looking for, is the vertice FARTHEST away from the origin.

 

There can be two vertices that fit this, but they will end up giving you the same answer.

 

So all you need to do is find the coordinates of where two lines intersect. (they should intersect at the said vertice)

 

 

I will leave that up to you to find this.

 Nov 19, 2019
edited by CalculatorUser  Nov 19, 2019
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Thanks, CU.  You not only showed the OP how to do it, you showed me how to do it, too.  If the wonderman had had his way, neither the OP nor I would have learned anything. 

 

Just to make sure I understand, you solve the first expression for y in terms of x, then plot a line as equal –3 and then plot a line as equal +3 and shade between the lines.  Then do the same thing with the other expression.  Is this right? 

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Guest Nov 19, 2019

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