# 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 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**

We have to solve for a, b and c from the given system of equations.

a +3b + 2c = 6…(1)

a – b + 4c = 2…(2)

c – b - a =2…(3)

Add (1) and (3)

=> a +3b + 2c + c – b – a =6 + 2

=> 2b +3c =8

Add (2) and (3)

=> a – b + 4c + c – b – a = 2+2

=> -2b +5c =4

Now by adding 2b +3c =8 and -2b +5c =4

We get 8c=12 => c= 12/8 = 3/2

Substitute c= 3/2 in 2b +3c =8

=> 2b +3*(3/2) = 8

=> 2b = 8- (9/2)

=> 2b = 7/2

=> b= 7/4

Now substitute b= 7/4 and c= 3/2 in c-b-a=2

=> 3/2 – 7/4 –a =2

=> a = 3/2 -7/4 -2

=> a = -9/4

**Therefore a = -9/4 , b= 7/4 and c= 3/2**

To solve the system of equations:

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

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

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

Solution:

(2)+(3) gives: 5c-2b =4...........(4)

(1)-(2) gives: 4b -2c = 4...........(5).

2*(4)+(5) gives: 10c-2c = 8+4 = 12. So 8c = 12 or c = 12/8 = 1.5

From (5) 4b-2c= 4. Or 4b-3 = 4. So 4b = 3+4=7. S b = 7/4 = 1.75.

Put b= 1.75 and c= 1.5 in (3). C-b-a = 2. Or 1.5-1.75-a = 2.

a = 1.5-1.75-2 = -2.25.

a = -2.25 , b = 1.75 and c= 1.5