Find the value of each variable from the following equations?

Question

A + C = 0
4A + B + D = 0
5A + 4B + 4C = 4
5B + 4D = 1

Answer 1

You need to try to get rid of some of the variables to make this thing easier to solve. For starters, let's number the equations:

(1) A + C = 0
(2) 4A + B + D = 0
(3) 5A + 4B + 4C = 4
(4) 5B + 4D = 1

OK, now notice that there are only 2 equations that have D in them, (2) and (4). Let's start there to get rid of D. Multiply (2) by 4 to get the D terms the same:
4(4A + B + D) = 0 * 4
16A + 4B + 4D = 0 (we'll call this eq. (5) )

Now we can subtract (4) from (5) to get rid of the D terms:
(5) 16A + 4B + 4D = 0
(4) - (0A + 5B + 4D = 1) <-- I added in the 0A to make things line up better
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(16 - 0)A + (4-5)B + (4-4)D = (0-1)
16A - B = -1 (we'll call this eq. (6) )

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OK, now notice that there are only 2 equations in the original set that have C in them, (1) and (3). Multiply (1) by 4 to get the C terms the same:
4(A + C) = 0 * 4
4A + 4C = 0 (we'll call this eq. (7) )

Now we can subtract (7) from (3) to get rid of the C terms:
(3) 5A + 4B + 4C = 4
(7) - (4A + 0B + 4C = 0) <-- I added in the 0B to make things line up better
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(5-4)A + (4-0)B + (4-4)C = (4-0)
A + 4B = 4 (we'll call this eq. (8) )

Now get A all by itself in eq. (8):
A + 4B = 4
A = 4 - 4B (we'll call this eq. (9) )

Now that we know what A is, we can plug this back into eq. (6) to solve for B!
16A - B = -1
16(4 - 4B) - B = -1
64 - 64B - B = -1
64 - 65B = -1
-65B = -65
B = 1

And now get A by plugging our B value back into eq. (9):
A = 4 - 4B
A = 4 - 4(1) = 4 - 4
A = 0

Now plug in the A value in eq. (1) to find C:
A + C = 0
0 + C = 0
C = 0

And plug in the B value in eq. (4) to find D:
5B + 4D = 1
5(1) + 4D = 1
5 + 4D = 1
4D = -4
D = -1

So final answer: A = 0, B = 1, C = 0, and D = -1.
You can plug these values back into the original equations as a check to prove to yourself that they work.

Answer 2

A + C = 0 <-- Eq 1
4A + B + D = 0 <-- Eq 2
5A + 4B + 4C = 4 <-- EQ 3
5B + 4D = 1 <-- Eq 4
Multiply Eq 2 by -4 and add result to Eq 4 getting:
-16A + B = 1 <-- Eq 5
Substitute C = -A (from Eq 1) into Eq 3 getting
A + 4B = 4 <-- Eq 6
Multiply Eq 6 by 16 and add to Eq 5 getting:
65B = 65
B=1
Substitute b+1 into Eq 6 getting:
A +4*1 =4 --> A = 0
Substitute A = 0 into Eq 1 getting: C= 0
Sustitute A=0 and B 1 into Eq 2 getting:
D= -1
So A=0, B= 1 C=0 and D= -1

Answer 3

A + C = 0-------------------(1)
4A + B + D = 0------------(2)
5A + 4B + 4C = 4---------(3)
5B + 4D = 1----------------(4)
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A + C = 0 => A = -C and vice versa
Substituting it in (3)
=>5A + 4B + 4(-A) = 4
=> A + 4B = 4----------------(5)

Take 4 x (2)
4A + B + D = 0 x4
=> 16A + 4B + 4D = 0
from (4) 4D = 1 - 5B
=> 16A + 4B + 1 - 5B = 0
=> 16A - B = -1-------------(6)

from (5) - 4x (6)
=> A + 4B = 4
(-) 16A - B = -1 x 4
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=> A - 64A + 4B -4B = 4 - 4
=> -63A = 0
=> A = 0
Sub in (5) = > B = 1
Sub in (4) => 4D = 1 - 5B
=> D = -1
As A = - C => C = 0

therefore,
A = C = 0
B = 1
D = -1

Peace out.