Solve the following systems of equations 2a+4b+c=5 ; a-4b=-6 ; 2b+c=7.

Mathematics

1 answer:

SOVA2 [1]2 years ago

8 0

Answer:

a = -2

b = 1

c = 5

Step-by-step explanation:

Given:

2a + 4b + c = 5 ............(1)

a - 4b = - 6

a = 4b - 6 .............(2)

2b + c = 7

c = 7 - 2b ...........(3)

substituting 2 and 3 in 1, we get

2(4b - 6 ) + 4b + (7 - 2b) = 5

8b - 12 + 4b + 7 - 2b = 5

10b - 5 = 5

b = 1

substituting b in 2, we get

a = 4(1) - 6

a = -2

substituting b in 3, we get

c = 7 - 2(1)

c = 5

thus,

a = -2

b = 1

c = 5

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