# System of linear equations

• Nov 11th 2009, 05:15 PM
dkpeppard
System of linear equations
I've worked through this one, but I must be doing something wrong because my answer is not coming out correctly. I need to solve this system of linear equations and will be using the addition method to eliminate one of the variables:

$y=-\frac{2}{3}x-3$

$x=-\frac{3}{2}y+9$

Now, I broke this down to the following, and this is where I think I am messing up. I presume I need to get 'x' and 'y' on the same left side, so I did:

$\frac{2}{3}x+\frac{1}{1}y=-3$

$\frac{1}{1}x+\frac{3}{2}y=9$

Is this correct, or am I missing something? I know in the end I should end up with a separate value for 'x' and 'y', but when I add the two equations together I end up getting something like (don't laugh) 0x+0y=54, which obviously is wrong, so I cannot finish the equation.

• Nov 11th 2009, 07:15 PM
dkpeppard
I further broke this down to:

2x + 3y = -9
and
2x + 3y = 18

In order for me to apply the addition method to this, I need to make the coefficient of one of the variables equal and opposite. I assume that means I multiply the first equation by 2 and the second by -2.

So I would get 2*2x+2*3y=2*-9, which equals 4x+6y=-18 for equation 1.

For equation two I would get -2*2x+-2*3y=-2*18, which equals -4x-6y=-36.

Using the addition method and adding equation one and two together, this all looks good for 'x', but my 'y' value also becomes zero, which leaves me no equation for the -54. I essentially get 0x + 0y = -54, which doesn't work.

Where am I going wrong here, please?
• Nov 11th 2009, 09:22 PM
Wilmer
You're doing ok; answer is simply "no solution".
• Nov 12th 2009, 07:37 AM
dkpeppard
Quote:

Originally Posted by Wilmer
You're doing ok; answer is simply "no solution".

Ok, thanks Wilmer. I literally worked this problem over and over for a couple hours, pulling my hair out. I thought I was doing it right, but because I am not very confident in math (I am back in school after twenty years to finish my degree), I was doubting myself.

Thank you again.
• Nov 12th 2009, 07:58 AM
Defunkt
Also, if you got it down to:

$(1) \ 2x + 3y = -9$

$(2) \ 2x + 3y = 18$

Then:

$(1) \ - \ (2) = 2x + 3y - (2x + 3y) = 2x + 3y -2x -3y = 0$
But also, $(1) \ - \ (2) = -9 - (18) = -9 -18 = -27$

So in total, you get $-27 = 0$, which is obviously a contradiction, and thus the system has no real solutions.
• Nov 12th 2009, 08:03 AM
dkpeppard
Quote:

Originally Posted by Defunkt
Also, if you got it down to:

$(1) \ 2x + 3y = -9$

$(2) \ 2x + 3y = 18$

Then:

$(1) \ - \ (2) = 2x + 3y - (2x + 3y) = 2x + 3y -2x -3y = 0$
But also, $(1) \ - \ (2) = -9 - (18) = -9 -18 = -27$

So in total, you get $-27 = 0$, which is obviously a contradiction, and thus the system has no real solutions.

Awesome, thanks for the further clarification.