ErinWeeks
  • ErinWeeks
Graph the solution set of the system of inequalities or indicate that the system has no solution.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ErinWeeks
  • ErinWeeks
x^2 + y2^ ≤ 49 5x + 4y ≤ 20 @Psymon could you help me with this?
Psymon
  • Psymon
Hmm, a circle and a line, lol.
ErinWeeks
  • ErinWeeks
i am not good with graphing at all!!

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Psymon
  • Psymon
Well, in this case our elimination method from before isn't much of an option. Probably impossible really. This means we must do substitution. The idea behind substitution is we pick one equation and then get either x or y in that equation all by itself. Just as if we were solving a normal algebra problem like 2x -5 + 13. We would get x by itself. Same thing with this. Now for the top equation, getting x or y by itself isn't practical. It doesn't do any good. So we want to use the bottom equation and see if we can solve for x or y. Which one we solve for does not matter. Since y has the smaller number multiplying it, I'll just solve for y in the second equation. So we have: 5x+4y <= 20 I first subtract 5x from both sides and then divide everything by 4 5x + 4y <= 20 -5x -5x 4y <= 20 -5x\[y=\frac{ 20-5x }{ 4 }\] So now we've solved for y. This result we then must substitute intot he other equation. Make sense so far?
ErinWeeks
  • ErinWeeks
Yea i think i got it so far, its making sense!
Psymon
  • Psymon
Okay, awesome. So here comes probably the more difficult part. We have to take what we just came up with and replace it for y in the other equation. Meaning we have this: \[x ^{2}+(\frac{ 20-5x }{ 4 })^{2}=49\]So yeah, pretty ugly here. But this is doable. Me personally, I think I would want to rewrite that fraction squared part. Be aware that the numerator is being squared and the denominator is being squared. What I'm going to do is ONLY square the bottom. Doing that will let me manipulate some things. So this is what I get: \[x ^{2}+\frac{ (20-5x)^{2} }{ 4^{2} } =49\] \[x ^{2}+\frac{ (20-5x)^{2} }{ 16 }=49\] So now that I have this, I can eliminate the fraction. Because I have a division of 16, I can eliminate it by a multiplication of 16, the opposite. This requires me to multiply everything by 16. That will give me: \[16x ^{2}+(20-5x)^{2}=(49)(16)\] Notice how I didn;t actually multiply 49 and 16, I just put them next to each other like a multiplication. The reason I do this is sometimes it is not necessary to multiply right away. We might be able to cancel something or simplify things easier by not multiplying. Anyway, if I multiply them and get 784, I can't even tell there's a 16 or a 49 in there. It's a lot harder to see the factors. Okay, so now I have to square that (20-5x) portion. There is a formula for a binomial squared that is: \[(a-b)^{2}=a ^{2}-2ab+b ^{2}\]Now I'll still multiply it out the normal foil way to show you, though. So I have (20-5x)(20-5x) So foil now 20*20 = 400 20*-5x = -100x -5x*20 = -100x -5x*-5x = 25x^2 combining them I get 25x^2 - 200x + 400. So now let's see what I have when I combine everything: 16x^2 + 25x^2 - 200x + 400=(49)(16) 41x^2 - 200x + 400-(49)(16) = 0 Now I move everything to one side of the equation because this is going to be a quadratic polynomial, meaning we have to factor it. Problem is the factoring of it looks like it'll be quite annoying. We can still try, though. Now, looks like I'm going tobe forced to multiply out that (49)(16), so I do that to get 784. SO 400-784 = -384, giving us 41x^2 - 200x - 384 = 0 Okay, this is ridiculous, haha. This is the process you'd have to do, though x_x I'm guessing the equation isn't expecting you to get the actual solutions, just get a general idea because I mean.......yeah, the answers for this are insane. But the idea is do you get what I did up to this point?
ErinWeeks
  • ErinWeeks
i got it so ar i probably wont remember everything but i got it so far.. This is a lot :0
Psymon
  • Psymon
Well, this isnt something that can easily be done by hand. Without a graphing calculator like I have, this requires calculus to get good results. But here's what we can do. We can still graph both of the equations and then get a good estimate of what our answers will be. Because these are inequalities, we don't have to be exact, we can be close and still get correct answers.
Psymon
  • Psymon
So, do you think you would know how to graph the bottom of the two equations?
ErinWeeks
  • ErinWeeks
i have know idea how to graph it .. Im sorry!! & I know i should have bout a graphing cal but i thought the one i bought had graphing but it doesnt and im half done with my pre calc so i thought i shouldnt get one.
Psymon
  • Psymon
I wouldnt get a graphing calculator unless youre going to docalculus and higher. Theres no point to it with anything less than that. But okay, do you remember y = mx + b?
ErinWeeks
  • ErinWeeks
a little but not a lot though.
Psymon
  • Psymon
Do you remember what m and b mean in that equation by chance?
ErinWeeks
  • ErinWeeks
b is the y-intercept i believe and i think m is the slope?
Psymon
  • Psymon
RIght. Hence why it is called slope-intercept form. The random m and b in the equation represent slope and intercept. The idea is when we have x and y to the first power, we can make the equation look like y = mx + b When we make it look like that, it is very easy to graph it. So given we have 5x + 4y <= 20, we now solve for y. Now we already did solve for y way up above and we got: \[y = \frac{ 20-5x }{ 4 }\] This isnt quite y = mx + b, but we only need a slight change. Now, every term in the numerator of a fraction can be made into its own separate fraction. For example: \[\frac{ x ^{3}-x ^{2}-x - 1 }{ 5 } \]Here I have a fraction with 4 terms in the numerator. This means I can make 4 separate fractions. All I do is take one of those terms and put it on top of our original denominator. So when i split this into 4 fractions, i get: \[\frac{ x ^{3} }{ 5 }-\frac{ x ^{2} }{ 5 }-\frac{ x }{ 5 }-\frac{ 1 }{ 5 } \] So what I just did for this example I can do for our problem. So: \[\frac{ 20-5x }{ 4 } \implies \frac{ 20 }{ 4 }-\frac{ 5x }{ 4 }\] \[y = 5 - \frac{ 5x }{ 4 } \] So now we have the form we want. Can you tell what the slope and what the y-intercept are?
ErinWeeks
  • ErinWeeks
is slope 5x? and y-intercept 4?
Psymon
  • Psymon
Nope. The y-intercept is the number not with x. So that means the y-intercept is +5. The slope is the number paired with x. Now the number with x in this case is NEGATIVE (5/4), We have to be wary of the sign, too. So what a -5/4 slope tells us is each point is down 5 and right 4 from the previous point. So we continuously go down 5 right 4 down 5 right 4, etc, etc. And our line crosses the y-axis at y = 5. So with that information, we can definitely graph this: |dw:1378451819219:dw| Kinda make sense?
ErinWeeks
  • ErinWeeks
kinda makes sense yea..
Psymon
  • Psymon
Alright, cool. So now we need to graph the circle. I dont suppose you have any idea about how to do that, do you?
ErinWeeks
  • ErinWeeks
nope i sure dont.. i feel bad i dont know how to do anything:/
Psymon
  • Psymon
Well, I still want to ask to be sure xD
ErinWeeks
  • ErinWeeks
i understand lol. if i knew it i would surely be happy to do it.
Psymon
  • Psymon
Okay, so the formula for a circle is this: \[(x-h)^{2}+(y-k)^{2}=r ^{2} \]Now in terms of remembering the formula, if you know pythagorean theorem, then you know: \[a ^{2}+b ^{2}=c ^{2} \]Well, thats exactly what a circle is! The only difference is the -h and the -k in there, otherwise its pretyt much a^2 + b^2 = c^2. In fact, it even means the same thing |dw:1378452287571:dw| The -h and -k parts indicate where the center of the circle is. The h is the x-coordinate and the kis the y-coordinate. Now we need to be careful. If I have: \[(x-2)^{2}+y ^{2}=16 \], this DOES NOT mean that h = -2 andour x-coordinate is -2. Even though it says minus, it's subtracting a positive number. The x-coordinate is actually positive 2. I If I have: \[x ^{2}+(y+3)^{2}=28 \]Now this time we see +3 but again, this does not mean that the k is a positive 3. Its actually -3. The reason is remember that I posted the formula with -h and -k. These have to be negative. But in this example I had a +3. Well, the only way thats possible is if its (y-(-3))^2 = (y+3). This is why the y-coordinate of the center of the circle is -3. Now also notice that your problem is just x^2 + y^2 = 49. There is no (x-5)^2 or anything like that. This means the center of the circle is the origin. A long explanation, especially since your circle is at the center, but we still need to know what to do if we got a problem that had this -h -k thing. Dont know if any of that made sense xD
ErinWeeks
  • ErinWeeks
it made sense but i am little confused. i think i got it though, might just have to re read it once more. lol
Psymon
  • Psymon
Yeah, of course. And its trying to explain it without tons and tons of examples, lol. But in the end, your circle has a center right at (0,0), so makes things easy. Now all of it = 49, which is r^2. So if we want to know the actual radius, we just take the square root of our number. So in this case, we have sqrt(49) = 7, meaning the radius of our circle is 7. A radius of 7 means that no matter which direction I face, the edge of the circle is always 7 units away fromthe circle. Graphing that I get:
Psymon
  • Psymon
|dw:1378453232743:dw| Im sorry, but I need to go here, though. Im literally passing out in the middle of typing this @_@
ErinWeeks
  • ErinWeeks
im sorry nightt !!! thankk you
Psymon
  • Psymon
No worries, not your fault xD After this, all you need to do is test points. Like, if you pick the point (0,0), is it true in both of your beginnign equations? So you plug in 0 for x and for y in each equation. If it makes BOTH true, you shade the area of the point you tested. So you test points inside and outside the lines of your graph until you have checked all the regions. But yeah, night xD

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