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Which ordered pair represents a solution to the following system of inequalities?

{2x+4y≤12

{3x−y<2

TandCforlife  Jun 3, 2017
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8+0 Answers

 #1
avatar+75376 
+1

 

Don't see any solution presented, but.....it will lie in the over-lapping shaded regions shown here :

 

 

 

cool cool cool

CPhill  Jun 3, 2017
 #2
avatar+4174 
+1

LOL !!!

 

Looks like the shaded regions are also absent today laughlaughlaugh

hectictar  Jun 3, 2017
 #3
avatar+75376 
+2

LOL!!!

 

Yep.....It would have helped to supply the link  :

 

https://www.desmos.com/calculator/2thymd2pwe

 

 

 

cool cool cool

CPhill  Jun 3, 2017
 #4
avatar+821 
+2

You can also solve this system algebraically, too, but it requires some logical thinking. You solve it similarly as that of a system of equations. Here are your inequalities:

 

1. \(2x+4y\leq12\)

2. \(3x-y<2\)

 

I'll use substitution. I'll solve for on both equations. I'll solve for on the inequality \(2x+4y\leq12\):

 

\(2x+4y\leq12\) Here is the original inequality. Subtract 2x first to begin isolating y.
\(4y\leq-2x+12\) Divide by 4 on both sides of the equation
\(y\leq-\frac{1}{2}x+3\) Finally, is isolated
   

 

Next, isolate in the inequality 2, \(3x-y<2\):

\(3x-y<2\) This is the original inequality. Subtract 3x on both sides
\(-y<-3x+2\) Divide by -1 on both sides. Remember that the inequality sign flips when dividing by a negative number!
\(y>3x-2\)  
   

 

Our inequalities have changed to from their original to being solved for y:

  1. \(y\leq-\frac{1}{2}x+3\)
  2. \(y>3x-2 \)

This is where the logical thinking comes to play.If 3x-2 is less than y and \(-\frac{1}{2}x+3\) is equal to and greater than y, then \(3x-2<-\frac{1}{2}+3\)! Now that there is one variable in this inequality, we can solve for x.

 

 

\(3x-2<-\frac{1}{2}x+3\) This is what we concluded above. Add 2 on both sides
\(3x<-\frac{1}{2}x+5\) Add (1/2)x on both sides
\(\frac{7}{2}x<5\) Multiply by 2/7 to isolate x.
\(x<\frac{10}{7}<1\frac{3}{7}\)  
   

 

Yet again, it requires a bit of logic again. Plug in 10/7 into both inequalities! Yes, you must plug it into both inequalities! You'll see why after the calculations are made. First, I'll plug into the first and second equation

 

\(y\leq\bf{-\frac{1}{2}*\frac{10}{7}}+3\) Simplify the right hand side by evaluating \(-\frac{1}{2}*\frac{10}{7}\)first.
\(y\leq-\frac{5}{7}+3\) Change 3 to an improper fraction so you can add the fractions together!
\(y\leq-\frac{5}{7}+\frac{21}{7}\) Add the fractions now because of the common denominators!
\(y\leq\frac{16}{7}\)  
   


Great! Let's plug in for the second equation and see what we get:

 

\(y>3(\frac{10}{7})-2\) Multiply 3 by 10/7 first
\(y>\frac{30}{7}-2\) Convert the 2 to an improper fraction
\(y>\frac{30}{7}-\frac{14}{7}\) Subtract the fractions to get the solution set for the other equation!
\(y>\frac{16}{7}\)  
   

 

Now that we know our x- and y-values, let's write them out as a solution set:

 

\(((x<\frac{10}{7},\frac{16}{7}

 

I guess you could stop here and move on to the next problem, but do you notice how y is really equal to every number? Currently, the solution set is saying that y is greater than 16/7, equal to 16/7 and less than 16/7. That's a verbose way of saying that y can be any of the real numbers. I'll write it out for you. Sorry, I had a hard time making the real-number symbol in LaTeX:
 

\((x\in{\rm I\!R}| x<\frac{16}{7},y\in{\rm I\!R})\)

 

You are done now because you have identified the possible values for x and y! You can verify this solution set by graphing. Luckily, Cphill has already done that! You can check it out, if you'd like.

 

One last note before you go!

 

I spent around 3 hours in total (with a slight respite in between) to generate this response. I truly attempted to provide a complete answer with comprehensible explanations. I hope it helped! Also, if anyone sees a typo, please tell me. I'm 95% confident that I missed one or two.

TheXSquaredFactor  Jun 3, 2017
edited by TheXSquaredFactor  Jun 3, 2017
edited by TheXSquaredFactor  Jun 3, 2017
edited by TheXSquaredFactor  Jun 3, 2017
edited by TheXSquaredFactor  Jun 3, 2017
edited by TheXSquaredFactor  Jun 3, 2017
 #5
avatar+821 
0

Okay, I can't seem to fix that typo, \((x<\frac{10}{7},\frac{16}{7}. I've tried and tried and tried, but it simply never saves my correction afterward. It's annoying, so annoying. Anyway, it's supposed to be:

 

 

(x<10/7,16/7 [insert less than symbol] y [insert less than and equal to] 16/7

TheXSquaredFactor  Jun 3, 2017
edited by TheXSquaredFactor  Jun 3, 2017
edited by TheXSquaredFactor  Jun 3, 2017
 #8
avatar+4174 
+1

Did you put this:     \)

at the end

 

Like this:

 

\(    x  <  \frac{10}{7}     ,     y  \leq  \frac{16}{7}    \)

 

???

 

backslash open and backslash close laugh

 

Sometimes it does do funny stuff though, especially if your problem takes up more than one line...... but the button normally works right. smiley

hectictar  Jun 4, 2017
 #6
avatar+75376 
0

 

Thanks for that detailed answer, X2...!!!

 

 

 

cool cool cool

CPhill  Jun 3, 2017
 #7
avatar+821 
+1

That's me... Mr. DetailedAnswer ... 

TheXSquaredFactor  Jun 3, 2017

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