7x1+2x2+3x3=15
5x1-3x2+2x3=15
10X1-11X2+5x3=36

9514 1404 393

  (x1, x2, x3) = (2, -1, 1)

Step-by-step explanation:

Your graphing calculator or an on-line solver can reduce the augmented matrix for you.

The result is the reduced row-echelon form ...

  (x1, x2, x3) = (2, -1, 1)

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Manual solution

As you probably know, the idea is to eliminate variables by using combinations of rows of coefficients. It is useful to start by creating a row with 1 in the first column. That can be done many ways, including ...

The first of these choices has the disadvantage of introducing fractions right away. The result of the 3rd choice is the row ...

  [1, -13, 0 | 15]

Now, the rest of column 1 can be made zeros by adding the row to the product of the first item in the row and this row. The row operations are ...

  r2 ← (r2 -5×r1) ÷ 2

  r3 ← r3 -10×r1

This gives ...

The row operation ...

  r2 ← 6×r3 -23×r2

will put a 1 on the diagonal of that row, making it be [0, 1, 7 | 6]. The rest of the second column can be made 0 by ...

  r1 ← 13×r2 +r1

  r3 ← (-119×r2 +r3)/(-828)

This gives ...

Finally, the replacements ...

  r1 ← r1 -91×r3

  r2 ← r2 -7×r3

will give the reduced row-echelon form shown in the top section of this explanation.

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Additional comment

For a 3-variable system of equations, it can work reasonably well to use available methods to solve the 2-variable system that results from substitution for the third variable. One of my favorite methods is to use a graphing calculator.

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

d. you cannot tell because there are no values shown.