anonymous
  • anonymous
Find the coordinates of the points of intersection of the line y + 2x = 11 and the curve xy=12. Help, please. Forced to do problems with little knowledge of the subject.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
y=11-2x put in xy=12 x(11-2x)=12 11x-2x^2=12 2x^2-11x+12=0 find x ,then corresponding values of y
anonymous
  • anonymous
So, I must foil then I will get the x values, right?
tkhunny
  • tkhunny
No and absolutely not. "FOIL" isn't a verb. It's not even a thing. Just solve the equation. You can also go the other way. xy=12 ==> y = 12/x y + 2x = 11 ==> (12/x) + 2x = 11 and solve for x.

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anonymous
  • anonymous
I got x = 2/3 and x = 4. but I don't really know how to find the coordinates. My teacher didn't teach us this, she just told us we had to do these problems.
tkhunny
  • tkhunny
The first cannot possibly be a correct solution, Lest's see how you solved either version you have been presented.
anonymous
  • anonymous
I solved what I got from the other dude, but didn't make much sense. (12/x) + 2x = 11 (12/4) + 2(4) = 11 3 + 8 = 11
tkhunny
  • tkhunny
Have you ever solved for both solutions of a quadratic equation, or just "guess and check"?
anonymous
  • anonymous
Yes, why do you ask?
tkhunny
  • tkhunny
I ask because that leads to both solutions, rather than just one. Factoring... 2x^2-11x+12=0 ==> (2x - ____)(x - 4) = 0
anonymous
  • anonymous
2x-3, I missed type, x = 3/2
tkhunny
  • tkhunny
Good. There's a good thing to know, there. That '2' out in front suggests that denominators of solutions can be '2', but never '3'. Okay, we have both x-coordinates. How shall we find the corresponding y-coordinates?
anonymous
  • anonymous
I am not sure, what should I start with? Is there a certain equation I must use?
tkhunny
  • tkhunny
Either original equation will do. Substitute the now-known values and solve.
anonymous
  • anonymous
Are the y values 8 and 3? (3/2, 8) (4, 3)
tkhunny
  • tkhunny
There you go. No more questioning.
anonymous
  • anonymous
Thank you so much, I was just confused about the curve and such. Didn't know if there was any additional equations or anything else that needed to be used.
tkhunny
  • tkhunny
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