1. help with solving linear systems (addition/elimination)

Hello all,

I understand that the goal is to eliminate one of the variables and not simply add to two equations.

In this example:

Solve the system:
-2x + y = 2
-x + 3y = -4

simplifies to:
-6x - 3y = -6
-x + 3y = -4

I understand that:
-3(-2x + y) = -3(2) = -6x - 3y = -6 by distribution.

But I don't understand why -3 was used and not simply 3?

In other examples in my textbook,
9(3x+4y) = 9(13)
4(5x-9y) = 4(6)

-9 is converted to 9.

So what are you supposed to use? How do you determine what sign the coefficient should keep? The book I'm using simply shows my examples but doesn't explain why the different signs are used.

Thanks.

2. Re: help with solving linear systems (addition/elimination)

The goal is to eliminate a variable. In the first equation, the coefficient for x is negative and y is positive. In the second equation, the coefficient for x is negative and y is positive. If you want to cancel x, you need x to be positive in one of the two equations. If you want to cancel y, you need y to be negative in one of the two equations.

3. Re: help with solving linear systems (addition/elimination)

Originally Posted by alexcordero
Hello all,

I understand that the goal is to eliminate one of the variables and not simply add to two equations.

In this example:

Solve the system:
-2x + y = 2
-x + 3y = -4

simplifies to:
-6x - 3y = -6
-x + 3y = -4

I understand that:
-3(-2x + y) = -3(2) = -6x - 3y = -6 by distribution.

But I don't understand why -3 was used and not simply 3?

In other examples in my textbook,
9(3x+4y) = 9(13)
4(5x-9y) = 4(6)

-9 is converted to 9.

So what are you supposed to use? How do you determine what sign the coefficient should keep? The book I'm using simply shows my examples but doesn't explain why the different signs are used.

Thanks.

-2x + y = 2 times -3 = 6x - 3y = -6. You show -6x - 3y = -6. First check if you made a typo or is the book wrong.

To answer your question, you're trying to eliminate the Y variable and solve for X. To do this, you need the coefficient of Y in the first equation to be the same as the coefficient of Y in the second equation but the opposite sign. In the second equation the coefficient of Y is +3 so you multiply the first equation by -3 to get the coefficient of Y to be -3.

simplifies to:
6x - 3y = -6
-x + 3y = -4

Notice when you add these two equations now you eliminate the Y variable and get 5X = -10. Solving for X you get X = -2. You can then substitute -2 for X in either equation to get the value of Y.

Steve

4. Re: help with solving linear systems (addition/elimination)

Originally Posted by SGS
-2x + y = 2 times -3 = 6x - 3y = -6. You show -6x - 3y = -6. First check if you made a typo or is the book wrong.
Thank you, Steve. Perceptive of you to note "my" typo, not the book. Sorry about that and thanks. It is in fact, 6x - 3y = -6

Alex