Solve the system of linear equations below.
2х + Зу = 2
х+ 6у = 4
А. x = 0, y=2/3
Bх = 2, y = -2/3
C. х = 4/3, у=2/9
D. x = 4, y = -2

Question

05-02-2020
Mathematics

contestada

Solve the system of linear equations below.
2х + Зу = 2
х+ 6у = 4
А. x = 0, y=2/3
Bх = 2, y = -2/3
C. х = 4/3, у=2/9
D. x = 4, y = -2

Respuesta :

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ItsDodo ItsDodo · Answer 1 · 2020-02-05T19:30:48+01:00

ItsDodo ItsDodo

05-02-2020

The answer is A

Multiply the 2nd equation by -2
-2x -12y = -8
Then add it to the 1st equation
-9y = -6
y = 2/3
By substituting in the 1st equation
2x + 3(2/3) = 2
2x = 0
x=0

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Solve the system of linear equations below. 2х + Зу = 2 х+ 6у = 4 А. x = 0, y=2/3 Bх = 2, y = -2/3 C. х = 4/3, у=2/9 D. x = 4, y = -2

Respuesta :

Otras preguntas

Solve the system of linear equations below.
2х + Зу = 2
х+ 6у = 4
А. x = 0, y=2/3
Bх = 2, y = -2/3
C. х = 4/3, у=2/9
D. x = 4, y = -2