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Need someone to check my work, and if wrong, tell me how....

1. How many solutions to this non-linear system -
y=(1/2)x^2+(x)-3
y=x-1

My answer is 2 solutions

2. Use the substitution method to solve this nonlinear system

y=x^2+2x+8
y=-6x-7

I got (11, -3) and (23, -5)

3. Use the elimination method to solve this nonlinear system
3x+y^2=19
x-y=7

I got (5, -2) and (6, -1)

4. How many REAL solutions to this nonlinear system

3x^2+2y=12
-3x^2+4y=24

I got 2 real solutions.

2006-07-23 17:06:51 · 6 answers · asked by metsfan6986 1 in Science & Mathematics Mathematics

6 answers

1)
Correct. There are two real solutions at (-2,-3) and (2,1)


2)
Close, but not quite....It's actually (-3,11) and (-5,23) [you had your x and y values reversed)


3)
Correct. Two real roots at (5, -2) and (6, -1)

4)
Wrong. Just one real solution, at (0,6)

2006-07-23 17:42:00 · answer #1 · answered by hfshaw 7 · 1 0

ok I will try to help you here

1. two solutions....did you get them? (2,1) and (-2,-3)

2. I think that you reversed the x and y in your solution pairs. (x,y) alpha order ! So the solutions are (-5,23) ans (-3,11) check these again in the original equations.

3. I assume that you did the correct method to find the answers..yours are correct. Did you run into a quadratic equation when you tried the elimination method? It factors so no big problem.

4. I am wondering what your two solutions are. If they are equal solutions then OK. (0,6) ? Some view this as one solution rather than 2 equal ones.

Enjoy your math !

2006-07-23 17:44:09 · answer #2 · answered by travlin 2 · 0 0

First problem: correct, the solutions are
(-2, -3) and (+2, +1)

Second problem: shouldn't you write the x up front, followed by the y? Otherwise correct,
(-3, 11) and (-5, 23)

Third problem: correct.

Last problem: I find only one solution, namely (x,y) = (0,6). Method: add the two equations together; the x's cancel, and you find 6y = 36, so y = 6. It follows immediately that 3x^2 = 0, so x = 0.

2006-07-23 17:39:36 · answer #3 · answered by dutch_prof 4 · 0 0

nr 4 ) 9x^2 = 0 => x = 0; one solution.

2006-07-23 17:43:12 · answer #4 · answered by gjmb1960 7 · 0 0

Numbers one and three are right. It looks like you might have reversed x and y for your answers in number two. I'd take another look at number four though.

2006-07-23 17:32:58 · answer #5 · answered by SmartGuy Dean 1 · 0 0

You should ask your teacher to provide fully worked solutions so that you can compare your work if your teacher is so lazy that he/she does not want to check it!

2006-07-24 03:06:13 · answer #6 · answered by Anonymous · 0 0

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