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

Question

Use echelon method to solve x and y in equations:
4x + y = 9
3x – y = 5
A) x = 1, y = 2
B) x = 2, y = 1
C) x = - 2, y = - 1
D) x = - 1, y = 2

Answer 1

4x + y = 9
3x - y = 5 Add them

7x = 14 solve for x

x=2
then solve for y

4(2)+y=9

8 + y = 9

y = 1

The answer isB

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

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Answer 4

B.