The cost of a car service is partly constant and partly varies with the time it takes to do the work. if cost 3500 naira for a 5 1/2 hour service and 2900 naira for a 4-hour service. Find the formula connecting cost, naira C with time, T hours, hence find the cost of a 7 1/2 hour service. find the formula connecting c and t using the elimination method
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C = r t + k
3500 = 5.5 r + k
2900 = 4 r + k
subtracting equations (to eliminate k) ... 600 = 1.5 r ... r = 400
substitute back to find k (the constant costs)
To find the formula connecting cost (C) with time (T) based on the given information, we can create a system of equations using the two given data points.
Let's represent the constant part of the cost with "a" and the part that varies with time with "b". We can set up the following equations:
Equation 1: 3500 = a + b * 5.5
Equation 2: 2900 = a + b * 4
To find the value of "a" and "b," we can use the elimination method:
Step 1: Multiply Equation 2 by 5.5 to make the coefficients of "b" in both equations equal:
5.5 * Equation 2: 5.5 * 2900 = 5.5 * a + 22b
Step 2: Subtract Equation 1 from the result of Step 1 to eliminate "a":
5.5 * 2900 - 3500 = 5.5 * a + 22b - (a + b * 5.5)
Now, simplify the equation:
15950 - 3500 = 4.5a + 16.5b
Simplify further:
12450 = 4.5a + 16.5b
This is our third equation.
We now have a system of three equations:
Equation 1: 3500 = a + b * 5.5
Equation 2: 2900 = a + b * 4
Equation 3: 12450 = 4.5a + 16.5b
Now, to solve this system of equations, you can use various methods such as substitution, elimination, or matrix operations. I will use the elimination method to find the values of "a" and "b".
Step 1: Multiply Equation 1 by 4 and Equation 2 by 5:
4 * Equation 1: 4 * (3500 = a + b * 5.5) -> 14000 = 4a + 22b
5 * Equation 2: 5 * (2900 = a + b * 4) -> 14500 = 5a + 20b
Step 2: Subtract Equation 2 from Equation 1:
14000 - 14500 = 4a + 22b - (5a + 20b)
-500 = -a + 2b
This is our fourth equation.
We now have a system of four equations:
Equation 1: 3500 = a + b * 5.5
Equation 2: 2900 = a + b * 4
Equation 3: 12450 = 4.5a + 16.5b
Equation 4: -500 = -a + 2b
To solve this system, you can use elimination again or use matrix operations. For simplicity, I will use the elimination method.
Step 1: Multiply Equation 4 by 4:
4 * Equation 4: 4 * (-500 = -a + 2b) -> -2000 = -4a + 8b
Step 2: Add Equation 3 to the result of Step 1:
-2000 + 12450 = -4a + 8b + (4.5a + 16.5b)
10450 = 0.5a + 24.5b
This is our fifth equation.
We now have a system of five equations:
Equation 1: 3500 = a + b * 5.5
Equation 2: 2900 = a + b * 4
Equation 3: 12450 = 4.5a + 16.5b
Equation 4: -500 = -a + 2b
Equation 5: 10450 = 0.5a + 24.5b
Using these equations, we can solve for "a" and "b" simultaneously. Once we have their values, we can write the formula connecting cost (C) with time (T) by plugging in the values of "a" and "b" into the formula:
C = a + bT
Finally, using the derived formula, you can substitute the value of T = 7.5 hours to find the cost of a 7 1/2 hour service.