+52 Which ordered pair represents a solution to the following system of inequalities?
{2x+4y≤12
{3x−y<2
+128474
Don't see any solution presented, but.....it will lie in the over-lapping shaded regions shown here :
+128474 LOL!!!
Yep.....It would have helped to supply the link :
https://www.desmos.com/calculator/2thymd2pwe
+2441 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 y on both equations. I'll solve for y 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, y is isolated |
Next, isolate y 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:
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 x 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 x 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.
+2441 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
