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using the addition (elimination) method. if the answer is a unique solution present it as an ordered pair (x.y) if not specify wether the answer is "no solution" or "many solutions"
3x - 2y = 7
-9x + 6y = -21

2007-02-24 10:56:12 · 7 answers · asked by Anonymous in Science & Mathematics Mathematics

7 answers

3(3x-2y=7) = 9x-6y=21
-9+6y=-21

answer is 0=0

you can tell that it is the same line because the answer is 0=0 (many solution)
if it was 5=3 or something like that, it means that it is parallel (no answer)

2007-02-24 11:02:14 · answer #1 · answered by ♫MIYA!! 2 · 0 0

OK... first you gotta set the equation to eliminate one of the variables. I will eliminate the y on top. To do this you must multiply the top equation by -3 in order to get the y to cancel out:

-3 (3x-2y=7) --------------- -9x+6y= - 21
-9x+6y=-21 --------------- -9x+6y= -21

Now once you do that you realize that the top and the bottom equation is the same which means that the solution is "many Solutions"

2007-02-24 19:06:19 · answer #2 · answered by Elizabeth M 3 · 0 0

Multiply 1st equation by 3 getting:
9x -6y =21
Add 2nd equation getting 0 = 0
Thus the two equations are the same lineand there are many solutions.

2007-02-24 19:03:25 · answer #3 · answered by ironduke8159 7 · 0 0

if you multiply the 1st equation by -3, you get the 2nd equation. so if you graphed them both, you'd see just 1 line, each of the infinitely many points on the line is a solution.

if the 2nd equation ended with = something other than -21, you'd have 2 parallel lines, no common points, no solution.

2007-02-24 19:00:56 · answer #4 · answered by Philo 7 · 0 0

(3x - 2y = 7 (3)
(-9x + 6y = -21

(9x - 6y = 21
(-9x + 6y = -21
-----------------
9x - 9x -6y + 6y = -21+21 = 0
no solutions
><

2007-02-24 19:06:18 · answer #5 · answered by aeiou 7 · 0 0

You honestly can't see that one equation is -3 times the other?? And that means no unique solution??


Doug

2007-02-24 19:01:54 · answer #6 · answered by doug_donaghue 7 · 0 1

many solutions

2007-02-24 19:05:13 · answer #7 · answered by Anonymous · 0 0

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