Use echelon method to solve x and y in equations:?

Question

Use echelon method to solve x and y in equations:
2x + 7y = - 8
- 2x + 3y = - 12
A) x = 3, y = 2
B) x = - 3, y = 2
C) x = 3, y = - 2
D) x = - 3, y = - 2

Answer 1

Hi,
The answer is x=3, y=-2.
Next time could you please say whether you want row-echelon form with back substation or reduced row-echelon form. I’ll use reduced row-echelon form because of the way you stated the problem. So, put the equations into an augmented matrix. (Coefficients and constants.)
Here’s the beginning matrix
[2…7..|..-8]
[-2...3.|..-12]…

[2…7..|..-8]
[0...10.|..-20]…R1+R2->R2 ..(This means add row #1 and row 2 and make the result the new row 2.)


[2…..0..|…6]…-7R2 +R1->R1
[0...... 1.|.-2]...(1/10)R2…(Do this step first.)

[1….0.|..3]…R1/2
[0....1.|.-2]

Answer: x=3, y=-2.

FE

Answer 2

Hi, The answer is x=2, y=1. But you say use the echelon method. There are two possible things that you might have in mind: ref (row-echelon form) or rref (reduced row-echelon form). I’ll do it using rref because of the way you stated the problem. So, put the equations into an augmented matrix. (Coefficients and constants.) Here’s the beginning matrix [4…1..|..9] [3...-1.|..5] [1…1/4..|..9/4]…1/4R1-->R1(This means multiply row #1 by ¼ [0......-7/4.|.-7/4]-3R1+R->R2 and make that the new row#1.) [1…1/4..|..9/4] [0......1.|…...1]…-4/7R2 This is now in row-echelon form. If you want this, then solve the equations using back substitution. [1…...0.|…..2]…-1/4R2 +R1-->R1 [0......1.|…...1 Answer: x=2, y=1. FE

Answer 3

The answer is C.
2x + 7y = -8
-2x + 3y = -12
Add the first equation to the 2nd equation.
10y = -20.
then y = -2
plug that in to the equation and you get x = 3.