Solve for A,B and C when
32A+14B+82C=664
11.5A+8B+52C=349
18A+26.2B-62C=560.4
Solution
(A,B,C) = (6,22,2)
To solve for the variables A, B, and C in the given system of equations, we can use the method of substitution or elimination. I'll use the elimination method to solve these equations step by step.
First, let's eliminate one variable by multiplying the equations by appropriate constants so that the coefficients of either A, B, or C will cancel out when we add or subtract the equations.
Looking at the coefficients of A in the equations, the most straightforward approach is to eliminate A by multiplying the second equation by 2 and the third equation by -1. We'll get:
1. (32A + 14B + 82C = 664)
2. (23A + 16B + 104C = 698)
3. (-18A - 26.2B + 62C = -560.4)
Now, let's eliminate the A variable by adding these equations together:
(32A + 14B + 82C) + (23A + 16B + 104C) + (-18A - 26.2B + 62C) = 664 + 698 - 560.4
Combine like terms:
32A + 23A - 18A + 14B + 16B - 26.2B + 82C + 104C + 62C = 664 + 698 - 560.4
37A + 3.8B + 248C = 801.6
Now we have one equation with only B and C variables left. Let's move on to eliminating another variable. Looking at the coefficients of B, we can eliminate B by multiplying the first equation by -4 and the second equation by 7. We'll get:
4. (-128A - 56B - 328C = -2656)
5. (80.5A + 56B + 364C = 2443)
Adding equation 4 and equation 5:
(-128A - 56B - 328C) + (80.5A + 56B + 364C) = -2656 + 2443
Combine like terms:
-47.5A + 36C = -213
Now, we have a system of two equations:
6. (37A + 3.8B + 248C = 801.6)
7. (-47.5A + 36C = -213)
Solving equations 6 and 7 can be done through elimination or substitution. I'll use substitution to solve for A and C.
From equation 7, we can express A in terms of C:
-47.5A = -36C - 213
A = (36C + 213) / 47.5
Substitute this value of A into equation 6:
37((36C + 213) / 47.5) + 3.8B + 248C = 801.6
Now, solve this equation for B and C.