Can anyone help me answer this simultaneous equation?
x+y=3 and y=x^2+x^-1
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M$8 Answers
Then you can solve for the other variable.
Boy... it's been a while. Hope that's right...
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M$3-x=x^2+x^-1;
multiplying both sides by x we get
3x-x^2=x^2+1;
2x^2-3x+1=0;
2x^2-2x-x+1=0
(2x-1)(x-1)=0
therefore
x=1 or x=1/2
now if x=1 then y=2
and if x=1/2 then y=5/2
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M$x + y = 3 becomes y = 3 - x
substitute
3 - x = x^2 + x^-1
0 = x^2 + x + x^-1 - 3
multiply both sides by x
0 = x^3 + x^2 - 3x + 1
x = 1 is a solution because 1^3 + 1^2 - 3(1) + 1 = 0
use polynomial division to find that (x^2 - 3x + 1) / (x-1) = x^2 + 2x - 1
use the quadratic formula to solve for the remaining roots, which are -1 +- 2^(1/2)
substitute back into the original equation x + y = 3
I graphed into just to make sure, and these are the correct answers
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M$The answer to your questions is that x=1 and y=2, x=4+sqrt2 y=-1-sqrt2, and x=4-sqrt2 y=-1+sqrt2
To do this, solve the first equation for y to get x=3-y. Then, substitute this into the second equation and get y=(3-y)^2+(3-y)^-1. Here simplify the equation to y=y^2-6y+9+1/(3-y). Move the y on the left to the right and the 1/(3-y) on the right to the left like so, -1/(3-y)=y^2-7y+9. Multiply everything by 3-y to simplify and get: -1=-y^3+10y^2-30y+27. Move the -1 over to the right and multiply by -1 to get 0=y^3-10y^2+30y-28. THen graph to find possible solutions and use the remainder theorem. On the graph you can see that it crosses at the point (2,0), so 2 is a solution. Use either long division or synthetic division to divide y^3-10y^2+30y-28 by x-2 (because 2 is a root). When you do that you get x^2-8x+14 as a remainder. Then use the quadratic equation to get the other two answers.h
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M$
Ah. I'm too detailed. Is that good or bad?
This is wrong. You say that (x-2) is a root of the polynomial y^3-10y^2+30y - 28. You should have said that (y-2) is a root. The correct answers are (1, 2) , (-1 - 2^(1/2), 4 + 2^(1/2)) , and (-1 + 2^(1/2), 4 - 2^(1/2)).
No, this is perfect. Thank you so much!
Whoops. Scratch the end part that says x-2 and replace it with y-2. Then you will get y^2-8y+14. So the answers are x=1 and y=2, x=-1+sqrt2 and y=4-sqrt2, and x=1-sqrt2 and y=4+sqrt2.
xy = x^3 + 1
Now substitute y = 3 - x in for y in the above equation
x(3 - x) = x^3 + 1
Expanding,
3x - x^2 = x^3 + 1
x^3 + x^2 - 3x + 1 = 0
There are a few ways to factor this. One would be to use the Cubic Formula directly, but that's not generally something you want to do unless you have no other choice. In this case, we can hope for rational roots and note that the rational roots theorem implies that the only possibilities are 1 and -1. Plugging in 1 shows that it is in fact a root, so we can just use long division to divide out the term (x - 1). This gives
(x - 1)(x^2 + 2x - 1) = 0
By the quadratic formula, the roots of the term on the right would be -1 +/- sqrt(2). Note that I didn't actually verify all the arithmetic there, but even if I made a mistake it illustrates how to solve such a problem.
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M$eq 1 becomes y=3-x
substitute 3-x for y in eq 2 gives
3-x=x^2+x-1 or
0=x^2+2x-4 which is a quadratic where a=1 b=2 and c=-4
the quadratic yields x=1.2361 and x=-3.2361 which is really
-1+5^1/2 and -1-5^1/2 5^1/2 is the square root 5)
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M$y=3-y
y=(3-y)(3-y) + 1/(3-y)
y= y*y - 6y+9 + 1/(3-y)
y-1/(3-y) = y*y-6y+9
3y-y*y-1=y*y-6y+9
0=2y*y-9y+10
0=(2y+1)(y-5)
y1=1/2, y2=5
x1=2.5, x2=2
X1=9 + sqrt(81-80)/2(2)=10/4=2.5
X2=9 - sqrt(81-80)/2(2)=8/4=2
The equation AX*X+BX+C=0 is quadratic
X=-b+/- sqrt(b*b-4ac)/2a
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M$
y - 1/(3-y)
multiply y by (3-y)/(3-y) to make a common denominator
(3-y)y/(3-y) - 1/(3-y)
subtract
((3-y)y - 1) / (3-y)
b/a - c/a = (b-c) / a
y-1/I(3-y)=0
multiple by (3-y)/(3-y) yields
(3-y)*y - 1/(3-y)*(3-y)
resulting in
-y*y+3y-1
This is wrong. When you went from
y-1/(3-y) = y*y-6y+9
to
3y-y*y-1=y*y-6y+9
you forgot the denominator when you subtracted. The left side of the equation should read 3y - y*y - 1 / (3-y)
-http://en.wikipedia.org/wiki/Ethics
-Dual Degree in Mathematics and Mechanical Engineering
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M$
In times like this, I've helped people by giving them tips on how to find the right answer rather then just giving the answer. Perhaps you could give some information on how to start the problem. (I would but I'm not that great in math.)
Why you have no evidence that it is and it is a single solution. The help provided for a single solution will hardly effect an overall grade. In addition this has to have been posted with enough time for the asker to enter it into homework and there would be no way to get the answer for an exam or quiz.
Besides with the number of incorrect solutions posted the asker is probably more confused no than before the question was asked.
Excellent way to answer!!
Aaah! Just tried to solve it myself. I can't remember how to solve a quadradic equation... I used to LOVE algebra!