Solve the system of equations using row operations. Write your answer in the form (x, y).

Question

{4y = 2x + 4

{-x = -2 - y

Answer 1

Start by getting all the variables on the left and the constants on the right in both equations:

{-2x+4y=4

{-x+y=-2

Since the equations are in this form now, you can put them in a matrix:

[-2 4 | 4]

[-1 1 | -2]

When working with a matrix, you want to get a 1 in the very first entry (top left), so switch the position on the equations:

[-1 1 | -2]

[-2 4 | 4]

Multiply the first row by -1 in order to make the first entry a positive 1:

[1 -1 | 2]

[-2 4 | 4]

In order to get a 0 below the top left 1, you need to multiply the top row by 2 and add it to the second row, so the first entry in the second row is 2*1 + (-2)=0, the second entry is 2*(-1) + 4=2, and the third entry is 2*2 + 4=8:

[1 -1 | 2]

[0 2 | 8]

After getting the zero in the first spot in the second row, you need to make the second entry in that row a 1, so divide that row by 2:

[1 -1 | 2]

[0 1 | 4]

The last step in the row reduction is to make the second entry in the first row a 0. Since the entries in the second spot are opposites, just add row 2 to row 1:

[1 0 | 6]

[0 1 | 4]

Now the row reductions are done! Translating back to equation form, we have x=6 and y=4, making your answer (6,4)

Checking your answer with the original equations, you see that 4*4=2*6+4, which gives 16=16, and -6=-2-4 gives -6=-6, so your answer satisfies both equations.

Hope this helped!

Answer 2

Row operations means you start with your equations in matrix form:

[ 2 4 4]

[ -1 1 -2]

You can add rows to each other, subtract rows from each other, and multiply rows by scalars.

I'm going to multiply the first row by 1/2, and then I'm also going to add it to the second row. I get this:

[ 1 2 2 ]

[ 0 3 0 ]

Now I'm going to multiply the second row by -1/3 and also add it to the first row, and I get this:

[ 1 0 2 ]

[ 0 1 0 ]

This gives us our solution: (x,y) = (2,0)