Question 515027
Let x represent the number of 4-wheel cars on the bridge, and let y represent the number of 6-wheel trucks.  We know that 32 vehicles were trapped on the bridge, so we can represent this fact algebraically as:

{{{ x + y = 32 }}}

Assuming that for every vehicle trapped, all tires on that vehicle had to be replaced, we know that 4x car tires had to be replaced, and 6y truck tires needed to be replaced.  The total number of tires that needed to be replaced was 148, and we can represent this fact algebraically as:

{{{ 4x + 6y = 148 }}}

Now we have a system of two linear equations in two unknowns.  There are a number of techniques to solve such a system; in this case, eliminating one variable will probably be easiest, since we can solve for x or y easily in the first equation.  We'll solve for y:

{{{ y = 32 - x}}} (subtracting y from both sides in the first equation)

So we now know that the number of trucks, y, must be 32 minus the number of cars, x.  We can take this equation and substitute for y in the second equation:

{{{ 4x + 6(32 - x) = 148 }}} (substituting 32 - x for y in the second equation)

Now we have an equation for x alone, so we solve:

{{{ 4x + 192 - 6x = 148 }}} (distributing the 6 on the left)
{{{ -2x + 192 = 148 }}} (combining like terms)
{{{ -2x = -44 }}} (subtracting 192 from both sides)
{{{ x = 22 }}} (divide both sides by 2)

So we got that x = 22.  Therefore we substitute back into our equation for y and get 

{{{ y = 32 - 22 = 10 }}}

So there were 22 cars and 10 trucks on the bridge.  Let's see if this makes sense.  22 cars and 10 trucks would make 32 vehicles total, so that works.  22 cars would mean 4(22) = 88 car tires, and 10 trucks would mean 6(10) = 60 truck tires, and that would mean 88 + 60 = 148 tires total, so that fact also works.  So we have the right answer: 22 cars and 10 trucks were on the bridge.