A truck is loaded with two different types of crates. When 20 of crate A and 25 of
crate B are loaded, the truck’s 8 tonne capacity is reached. When 2 of crate A and 16
of crate B are loaded, the same capacity is reached. Calculate the mass of each crate. Simultaneous equation please
20A + 25B = 8
2A + 16B = 8 ---> A + 8B = 4 ----> A = 4-8B
Use substitution ....
To solve this problem using simultaneous equations, let's denote the mass of crate A as 'a' (in tonnes) and the mass of crate B as 'b' (in tonnes).
According to the given information, when 20 crates of type A and 25 crates of type B are loaded, the truck's capacity of 8 tonnes is reached. So, we can set up the equation:
20a + 25b = 8
Similarly, when 2 crates of type A and 16 crates of type B are loaded, the same capacity of 8 tonnes is reached. We can set up the second equation:
2a + 16b = 8
Now we have a system of two equations:
20a + 25b = 8 ...(Equation 1)
2a + 16b = 8 ...(Equation 2)
To solve this system of equations, we can use any method such as substitution or elimination. I will use the elimination method:
First, let's multiply Equation 2 by 5 to make the coefficients of 'b' equal in both equations:
10a + 80b = 40 ...(Equation 3)
Now we can subtract Equation 3 from Equation 1 to eliminate 'b':
(20a + 25b) - (10a + 80b) = 8 - 40
20a + 25b - 10a - 80b = -32
10a - 55b = -32 ...(Equation 4)
Now we have two equations to solve:
10a - 55b = -32 ...(Equation 4)
2a + 16b = 8 ...(Equation 2)
We can multiply Equation 4 by 2 to simplify the system:
20a - 110b = -64 ...(Equation 5)
2a + 16b = 8 ...(Equation 2)
Now we can subtract Equation 5 from Equation 2 to eliminate 'a':
(2a + 16b) - (20a - 110b) = 8 - (-64)
2a + 16b - 20a + 110b = 8 + 64
-18a + 126b = 72 ...(Equation 6)
We now have two equations:
-18a + 126b = 72 ...(Equation 6)
2a + 16b = 8 ...(Equation 2)
Let's multiply Equation 6 by 9 to simplify the system:
-162a + 1134b = 648 ...(Equation 7)
2a + 16b = 8 ...(Equation 2)
Now we can add Equation 7 to Equation 2 to eliminate 'b':
(-162a + 1134b) + (2a + 16b) = 648 + 8
-160a + 1150b = 656 ...(Equation 8)
Now we have two equations:
-160a + 1150b = 656 ...(Equation 8)
2a + 16b = 8 ...(Equation 2)
Let's multiply Equation 8 by 5 to simplify the system:
-800a + 5750b = 3280 ...(Equation 9)
2a + 16b = 8 ...(Equation 2)
Now we can subtract Equation 2 from Equation 9 to eliminate 'b':
(-800a + 5750b) - (2a + 16b) = 3280 - 8
-800a + 5750b - 2a - 16b = 3272
-802a + 5734b = 3272 ...(Equation 10)
Now we have two equations:
-802a + 5734b = 3272 ...(Equation 10)
2a + 16b = 8 ...(Equation 2)
Let's multiply Equation 10 by 1/2 to simplify the system:
-401a + 2867b = 1636 ...(Equation 11)
2a + 16b = 8 ...(Equation 2)
Now we can add Equation 11 to Equation 2 to eliminate 'a':
(-401a + 2867b) + (2a + 16b) = 1636 + 8
-399a + 2883b = 1644 ...(Equation 12)
Now we have two equations:
-399a + 2883b = 1644 ...(Equation 12)
2a + 16b = 8 ...(Equation 2)
We can solve this system of equations to find the values of 'a' and 'b' by using the method of substitution or any other preferred method.