Posted by Completely lost?! on Thursday, September 10, 2009 at 10:31pm.
Do you see anything here:
7x+3y= pi 4x-6y= pi^2 2x+3y= 0 4x+6y= 0 2x+3y=1 4x+ 6y= 1
x+y=5 x+2y=10
I don't see anything wrong with the following:
x+y=5 x+2y=10
Do you see anything here?
2x-3y=5 4x-6y=10
I'm soryy i think i wrote this wrong there are like 4 questions in one of the questions
1. 7x+3y= pi 4x-6y= pi^2
2. 2x+3y= 0 4x+6y= 0 2
3. x+3y=1 4x+ 6y= 1
4. x+y=5 x+2y=10
5. 2x-3y=5 4x-6y=10
I am pretty sure i can figure out 2-4 (i wanted to check my answers on those though) but i am not sure how to do the fist one
The best way to check is to eliminate one of the two variables.
1. case U
If it comes up with ax+by=c like any other equation, there is a unique solution.
2. Case I
If it comes up with 0x+0y=0, this means that one equation is the linear transformation of the other, such as
3x+2y=4
6x+4y=8
Then we conclude that there is only one equation for two unknowns, the solution of which is
x=(4-2y)/3
Any value of y with the corresponding value of x is acceptable as solution, so this is case I, infinitely many solutions.
3. case N
If, after elimination of one variable, we come up with 0x+0y=5, which is impossible, then we conclude that there is no solution. Example:
x+y=3
2x+2y=8
Post your answers for a check if you wish.
1. case U
If it comes up with x=2, or 4y=6 like any other solution, there is a unique solution.
From http://www.jiskha.com/display.cgi?id=1252725501
1. 7x+3y= pi 4x-6y= pi^2
2. 2x+3y= 0 4x+6y= 0
3. x+3y=1 4x+ 6y= 1
4. x+y=5 x+2y=10
5. 2x-3y=5 4x-6y=10
1.U 2.N 3.N 4.U 5.I
Here is the check requested:
1. correct
2 If the equations are
2x+3y= 0
4x+6y= 0
as you posted above, the answer of No Solution is not correct. Try to eliminate one variable and see to which case it belongs.
3. Try again to eliminate one of the variables and see to which case it belongs.
4. correct
5. correct