# Issues with a Linear Equation

#### KH%3

##### New member
Hello fellow Mathers

( Intmath: Linear Equations: Example 2 (Link) )
I'm trying to figure out how exactly a specific equation works, where the guide tells me to do this example
3x + y = 10 [1]
x − 2y = 1 [2]
Explanation given:
If we subtract one row from the other row, we don't eliminate anything. However, if we multiply one of the rows, we can eliminate one of the variables by adding the rows.

Row [1] ×2 gives us
6x + 2y = 20 [3]
x − 2y = 1 [2] (no change)
If we add lines [3] and [2], we eliminate y.
So x=3 and using line [1], y=1.
Check in line [2]: 3−2(1)=1 [OK]
So our solution is (3,1).

I don't understand why we multiply it. I see that if i try calculating it, i get an imaginary ( 2.714...) number. But if I double it, that doesn't change. Perhaps I also made a mistake in the calculation, but could not find it yet. (it's a simple one, you should be able to replicate in less than 1 minute).

#### KH%3

##### New member
I may have figured this one out myself now. If you can, it's easyer to "eliminate" a variable, to get a result. Which seems to work in the other examples too. I wasn't sure what eliminate ment, until i examined this question a bit further. sorry for wasting your time.

#### Dr.Peterson

##### Elite Member
Yes, the goal of the method is to reduce the system of equations from two equations with two variables, to one equation with one variable, which you know how to solve. "Elimination" means removing a variable, making the system simpler. If that doesn't happen, then you haven't accomplished anything.

Can you explain what you meant by "if i try calculating it, i get an imaginary ( 2.714...) number "? That number isn't imaginary, and I can't see where it would come from. Possibly there is another misunderstanding we can correct.

#### HallsofIvy

##### Elite Member
There are many different ways to solve systems of linear equations. With 3x+ y= 10 and x- 2y= 1, seeing the "+ y" and "-2y" we see we can get the same thing, with opposite signs, in the first equation and them adding the two equations "eliminating" the y terms, leaving a single equation in x.

But we could multiply the second equation by -3, giving us 3x+ y= 10 and -3x+ 6y= -3. Now adding the two equation "eliminates" the x term leaving 7y= 7 so that y= 1. Then x- 2y= x- 2= 1 so x= 3. That is, x= 3 and y= 1 as before.

A third method would to subtract 3x from both sides of the first equation: y= 10- 3x. Since y= 10- 3x, we can replace the y in the second equation to get x- 2y= x- 2(10- 3x)= x- 20+ 6x= 7x- 20= 1. Then 7x= 21 so x= 3 and y= 10- 3(3)= 10- 9= 1 as before.

Yet a fourth method would be to add 2y to both sides of the second equation; x= 1+ 2y. Now replace x in the first equation by that: 3(1+ 2y)+ y= 3+ 7y= 10. Subtract 3 from both sides to get 7y= 7 and then, again, 7y= 7 so y= 1 and x= 3.