I've set a problem up, something like this.
s^4: A+D=0
s^3: -2A+B-3D+E=0
s^2: 2A-B+C+3D-3E=0
s^1: -2A+B-D+3E=4
s^0: A+C-E=4
I'm sorry. I forgot to say that I don't know how to solve for A, B, C, D, and E... Help there would be appreciated.
I would let my calculator do the gaussian reduction.
It's easier to substitute for s the values that make the denominators zero and then demand equalty. You then get simpler equations than comparing the coefficients of equal powers of s on both sides.
Would anybody mind showing it out? I tried what Count Iblis said, but it didn't really work out well...
To solve the system of equations you've provided, you can use the method of elimination or substitution. I will guide you through the process of using elimination to find the values of A, B, C, D, and E.
Step 1: Start by multiplying the equations to create a system that only involves A, B, C, D, and E:
Equation 1: s^4: A + D = 0
Equation 2: s^3: -2A + B - 3D + E = 0
Equation 3: s^2: 2A - B + C + 3D - 3E = 0
Equation 4: s^1: -2A + B - D + 3E = 4
Equation 5: s^0: A + C - E = 4
Step 2: Use elimination to eliminate one variable at a time. Since all equations have A, let's start by eliminating A.
To eliminate A from Equation 2 and Equation 3, multiply Equation 2 by 2 and Equation 3 by -2:
2 * Equation 2: -4A + 2B - 6D + 2E = 0
-2 * Equation 3: -4A + 2B - 2C - 6D + 6E = 0
Adding these two equations together, you will eliminate A:
(-4A + 2B - 6D + 2E) + (-4A + 2B - 2C - 6D + 6E) = 0
Simplifying, you get:
-8A + 4B - 8D + 8E - 2C = 0 --> 4B - 8D + 8E - 2C = 8A
The new equation will be referred to as Equation 6:
Equation 6: 4B - 8D + 8E - 2C = 8A
Now we have eliminated A.
Step 3: Use elimination again to eliminate another variable. Let's eliminate B next.
To eliminate B from Equation 6 and Equation 4, multiply Equation 6 by -2 and Equation 4 by 4:
-2 * Equation 6: -8B + 16D - 16E + 4C = -16A
4 * Equation 4: -8A + 4B - 4D + 12E = 16
Adding these two equations together, you will eliminate B:
(-8B + 16D - 16E + 4C) + (-8A + 4B - 4D + 12E) = 16
Simplifying, you get:
-26E + 4C - 12A + 4D = 16 --> 4C - 12A + 4D - 26E = 16
The new equation will be referred to as Equation 7:
Equation 7: 4C - 12A + 4D - 26E = 16
Now we have eliminated B.
Step 4: Continue using elimination to eliminate another variable. Let's eliminate C next.
To eliminate C from Equation 7 and Equation 5, multiply Equation 7 by 2 and Equation 5 by -4:
2 * Equation 7: 8C - 24A + 8D - 52E = 32
-4 * Equation 5: -4A - 4C + 4E = -16
Adding these two equations together, you will eliminate C:
(8C - 24A + 8D - 52E) + (-4A - 4C + 4E) = 32
Simplifying, you get:
-28A + 8D - 48E = 16 --> -7A + 2D - 12E = 4
The new equation will be referred to as Equation 8:
Equation 8: -7A + 2D - 12E = 4
Now we have eliminated C.
Step 5: Use elimination again to eliminate another variable. Let's eliminate E next.
To eliminate E from Equation 8 and Equation 1, multiply Equation 8 by 4 and Equation 1 by 12:
4 * Equation 8: -28A + 8D - 48E = 16
12 * Equation 1: 12A + 12D = 0
Adding these two equations together, you will eliminate E:
(-28A + 8D - 48E) + (12A + 12D) = 16
Simplifying, you get:
-16A + 20D - 48E = 16 --> -4A + 5D - 12E = 4
The new equation will be referred to as Equation 9:
Equation 9: -4A + 5D - 12E = 4
Now we have eliminated E.
Step 6: Now, we have a system of three equations involving A, D, and E:
Equation 6: 4B - 8D + 8E - 2C = 8A
Equation 9: -4A + 5D - 12E = 4
Equation 5: A + C - E = 4
You can solve this system of equations using various methods, such as substitution or matrix algebra, to find the values of A, D, and E.