Answer: Let's assume that the trucking firm purchases x trucks of model A, y trucks of model B, and z trucks of model C.
We know that the total number of trucks purchased should be 10, so:
x + y + z = 10
We also know that the total additional shipping capacity provided by the trucks should be 28 tons, so:
2x + 3y + 5z = 28
Now we have two equations with three variables. We can solve for one variable in terms of the other two, and substitute that expression into the other equation to get an equation with only two variables. For example, we can solve the first equation for x:
x = 10 - y - z
And substitute into the second equation:
2(10 - y - z) + 3y + 5z = 28
Expanding and simplifying:
20 - 2y - 2z + 3y + 5z = 28
Combining like terms:
y + 3z = 4
Now we have two equations with two variables:
x + y + z = 10
y + 3z = 4
We can solve for y in the second equation:
y = 4 - 3z
And substitute into the first equation:
x + (4 - 3z) + z = 10
Simplifying:
x + 1z = 6
x = 6 - z
Now we have three equations with three variables:
x + y + z = 10
2x + 3y + 5z = 28
x = 6 - z
We can substitute the expression for x into the first equation:
(6 - z) + y + z = 10
Simplifying:
y = 4 - (6 - z)
y = z - 2
Now we have expressed all three variables in terms of z. We can substitute these expressions into the second equation and solve for z:
2(6 - z) + 3(z - 2) + 5z = 28
Simplifying:
12 - 2z + 3z - 6 + 5z = 28
Combining like terms:
6z + 6 = 28
Solving for z:
z = 3
Now we can use the expressions for x and y to find how many trucks of each model the company should purchase:
x = 6 - z = 3
y = z - 2 = 1
Therefore, the company should purchase 3 trucks of model A, 1 truck of model B, and 6 trucks of model C to provide exactly 28 tons of additional shipping capacity.
Explanation: