Solve a+3b+2c= 6, a-b+4c= 2 and c-b-a =2

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a + 3b + 2c = 6 .........(1)

a - b + 4c = 2.............(2)

-a - b + c = 2 .............(3)

To solve the system, we are going to use th elimination methos.

First we will add (2) and (3):

==> -2b + 5c = 4 ...........(4)

Now...

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a + 3b + 2c = 6 .........(1)

a - b + 4c = 2.............(2)

-a - b + c = 2 .............(3)

To solve the system, we are going to use th elimination methos.

First we will add (2) and (3):

==> -2b + 5c = 4 ...........(4)

Now subtract (2) from (1)

==> 4b - 2c = 4 ..........(5)

Now multiply (4) by (2) and add tp (5):

==> 8c = 12

==> c= 12/8

==> c = 3/2

Now to calculate b, we will substitute in (4)

-2b + 5c = 4

==> b= (5c - 4)/2 = 5*3/2 - 4 )/2 = 7/4

==> b = 7/4

Now to find a, we will substitute in (2)

==> a -b + 4c = 2

==> a= b - 4c + 2

            = 7/4 - 4*3/2 + 2

              = (7- 24 + 8)/4 = -9/4

==> a= -9/4

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