4a+2b+c=10
100a+10b+c=12
16a+4b+c=12
I need help to solve this problem. I keeping getting A=1.70 B=2.0 and C=5. My book says that is wrong, could you show me how to get the correct answer?
If we call the equations #1,2,3 then subtract #1 from #2 and #3 from #3 to get
96a + 8b = 2
84a + 6b = 0
So, 6b = -84a, meaning 8b = -112a
So,
96a - 112a = 2
-16a = 2
a = -1/8
now use that to get
b = 7/4
c = 7
It would have been nice for you to have shown your work. Oh well, maybe you can use the above logic to discover your error.
To solve this system of equations, we can use the method of elimination or substitution. Let's use the elimination method in this case.
First, we want to eliminate one variable in two of the equations. We can do this by multiplying the equations by appropriate constants to make the coefficients of one variable in two equations equal.
Let's start by eliminating 'a.' Multiply the first equation by 25, the second equation by 2, and the third equation by 5. This will make the coefficient of 'a' the same in the first and third equation:
100a + 50b + 25c = 250
200a + 20b + 2c = 24
80a + 20b + 5c = 60
Now, subtract the second equation from the first equation and the third equation from the first equation:
100a + 50b + 25c - (200a + 20b + 2c) = 250 - 24
100a + 50b + 25c - 200a - 20b - 2c = 226
-100a + 30b + 23c = 226 ........(Equation 4)
100a + 50b + 25c - (80a + 20b + 5c) = 250 - 60
100a + 50b + 25c - 80a - 20b - 5c = 190
20a + 30b + 20c = 190 ........(Equation 5)
Now, we have two equations (Equation 4 and Equation 5) in two variables 'b' and 'c.' We can solve this system of equations using the same technique.
Multiply Equation 4 by 2 and Equation 5 by 5 to eliminate 'b':
-200a + 60b + 46c = 452 ........(Equation 6)
100a + 150b + 100c = 950 ........(Equation 7)
Now subtract Equation 6 from Equation 7:
100a + 150b + 100c - (-200a + 60b + 46c) = 950 - 452
100a + 150b + 100c + 200a - 60b - 46c = 498
300a + 90b + 54c = 498 ........(Equation 8)
Now we have two equations (Equation 5 and Equation 8) with two variables 'b' and 'c.' We can solve this system of equations:
20a + 30b + 20c = 190 ........(Equation 5)
300a + 90b + 54c = 498 ........(Equation 8)
To solve this system, subtract 10 times Equation 5 from Equation 8:
300a + 90b + 54c - (20a + 30b + 20c) = 498 - 10(190)
300a + 90b + 54c - 20a - 30b - 20c = 498 - 1900
280a + 60b + 34c = -1402 ........(Equation 9)
Now we have one equation (Equation 9) with two variables 'a' and 'c.' We can solve this equation:
280a + 60b + 34c = -1402 ........(Equation 9)
Let's assume 'a' = 1, and substitute it into Equation 9:
280(1) + 60b + 34c = -1402
280 + 60b + 34c = -1402
60b + 34c = -1402 - 280
60b + 34c = -1682 ........(Equation 10)
Now we have one equation (Equation 10) with two variables 'b' and 'c.' We can solve this equation:
60b + 34c = -1682 ........(Equation 10)
Now, we need another equation to solve the system fully and find the values of 'a,' 'b,' and 'c.' Please provide another equation or the intended value of one of the variables if available.