Final answer:
The system of linear equations is solved using substitution, yielding the final solution for a, b, and c. By simplifying the equations progressively, we can find the numerical values for these variables i.e. a = 93/110 + 6, b = 31/110, c = 8 - 620/110.
Step-by-step explanation:
To solve the system of linear equations provided, we can use substitution or elimination methods. The equations are:
a + 4b + 6c = 23
3a + 2b + c = 26
6b + 20 = a + 14
First, we can solve the third equation for a:
a = 6b + 20 - 14
a = 6b + 6
Now we can substitute the expression for a into the first two equations:
(6b + 6) + 4b + 6c = 23
3(6b + 6) + 2b + c = 26
Simplifying these we get:
10b + 6c = 17
18b + 18 + 2b + c = 26
From which we get:
10b + 6c = 17 (Equation 4)
20b + c = 8 (Equation 5)
Now, solve for c from Equation 5:
c = 8 - 20b
And substitute it into Equation 4:
10b + 6(8 - 20b) = 17
Solve for b:
10b + 48 - 120b = 17
-110b = -31
b = 31/110
Now, we can find values of a and c using the expressions for a and c we derived earlier:
a = 6(31/110) + 6
c = 8 - 20(31/110)
After simplification:
a = 93/110 + 6
c = 8 - 620/110
Final solution: a = 93/110 + 6, b = 31/110, c = 8 - 620/110.
Given the system of linear equations:
a + 4b + 6c = 23
a + 2b + c = 2
6b + 20 = a + 14
Calculate values of a, b, and c?