# Help with augmented matrix methods, 3x1 + 4x2 = 1, x1 - 2x2 = 7?

Ok i'm not sure how to type this symbol, the numbers on the right of "x" are small numbers that are located bottom right of the x

3x1 + 4x2 = 1

x1 - 2x2 = 7

Can you please explain the steps, i am still not familiar with augmented matrix methods yet so any help is much appreciated!

Relevance

I'm assuming you have to row reduce that system that you wrote. I'm going to write it as

3x + 4y =1, x - 2y = 7, and then we won't have to worry about the subscripts you mentioned.

So let's put that system into a matrix

3 4 1

1 -2 7

I'm going to interchange rows, because I like to work with a 1 in the top corner.

1 -2 7

3 4 1

Now multiply by -3 on the top row, add it to the bottom row, and replace the bottom row.

1 -2 7

0 10 -20

Let's divide the bottom row by 10.

1 -2 7

0 1 -2

And that's it, yes? The bottom row gives us that 1*y = -2, or y = -2. The top row says 1*x - 2*y = 7, or that x -2(-2) = 7. So x = 7 - 4 = 3.

• First, get x and y intercepts of the restraints (the word "subject to" indicates these are restraints) a) X1 + 2X2 <=323 X1 + 2X2 = 323 if X2 = 0, then X1 = 32 ---- (32,0) if X1 = 0, then X2 = 16 ---- (0,16) b) 3X1 + 4X2 <= 84 3X1 + 4X2 = 84 if X1 = 0, then X2 = 21 ----- (0,21) if X2 = 0, then X1 = 28 ----- (28,0) c) X1, X2 >= 0 (non-negative restraints) ----- (0,0) X1 = x X2 = y then, plot the points on an x, y plane.... look for the shaded region...by the use of testing points (a point anywhere on the plane not on the line) then look for the corner points of the shaded region and substitute them to the objective (P = 50x + 80y) The highest P you can obtain is the maximum.

• umm my answer is different than the previous one

so ill explain mine

my aug.matrix was

3 4 | 1

1 -2 | 7

divide first row by 3

1 4/3 | 1/3

1 -2 | 7

R2 = r1-r2

1 4/3 | 1/3

0 -2/3 | -20/3

r2 = r2/(-2/3)

1 4/3 | 1/3

0 1 | 10

r1 = (-4/3*r2) + r1

1 0 | -13

0 1 | 10

so x1 = -13

and x2 = 10