# linear system

• August 8th 2008, 05:24 AM
euclid2
linear system
without graphing or using any algebraic methods determine whether or not there is a solution to the following

3x+y=2
6x-2y=3
• August 8th 2008, 05:46 AM
TKHunny
That is an utterly silly request. What's left, Astrology?

Easiest determination, in my view 3(-2) - 6(1) = -6 - 6 = -12 this is not zero (0). There is a unique solution. However, addition and subtraction seem awfully "algebraic" to me.
• August 8th 2008, 05:51 AM
earboth
Quote:

Originally Posted by euclid2
without graphing or using any algebraic methods determine whether or not there is a solution to the following

3x+y=2
6x-2y=3

If you have asystem of linear equations like:

$\left|\begin{array}{l} a_1 x+b_1 y = c_1 \\a_2 x + b_2 y = c_2\end{array}\right.$

then there exist a unique solution if the determinant

$D=\left|\begin{array}{cc}a_1 & b_1 \\a_2 & b_2 \end{array} \right| \neq 0$

With your example $D = -12 \neq 0$ and therefore there must be an unique solution.
• August 8th 2008, 05:54 AM
Dubulus
3x+y=2
6x-2y=3

3x + y would be half of 6x-2y except for the operator. You are left with 6x=4 and 6x=3. Again, this does use algebraic concepts, but I just did it without the equations.
• August 8th 2008, 06:05 AM
euclid2
i was thinking if you went
(3x+y=2)2
6x+2y=4
6x-2y=3
Since the X are the same when multiplied by two i thought there would be no solution
• August 8th 2008, 06:44 AM
Dubulus
thats close
but if you end up with

6x+2y=4
6x-2y=3

then one of the 6x would have to be -6x for them to cancel