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To solve problems related to motion, we need to understand the concept of distance. The distance formula helps calculate how far an object travels over a certain period. It is given by the equation: \[ \text{distance} = \text{speed} \times \text{time} \] This formula is essential, as it links the speed of an object to the time taken and the distance covered. For example, if a car travels at 60 km/h for 2 hours, the distance it covers is 120 km (since 60 km/h \times 2 hours = 120 km). Understanding this formula makes it easier to solve problems involving motion in different directions, like in the given exercise where two cars are moving in opposite directions.

In algebra word problems involving motion, speed and distance are two key concepts. Speed is how fast an object is moving, typically measured in units like kilometers per hour (kph). Distance, on the other hand, measures how far the object has traveled. Let's break it down: When two objects move towards or away from each other, their speeds can be combined or subtracted to find the total speed. In our problem, two cars travel in opposite directions. Their combined speed becomes the sum of their individual speeds. If one car is traveling 20 kph and the other 35 kph, the total speed is 20 + 35 = 55 kph. Given the total distance and the time traveled, applying the distance formula helps determine unknown variables, such as the speed of each car.

Solving equations is a fundamental skill in algebra. It involves finding the value of a variable that makes the equation true. Let's see how it applies in our example: We start with the equation derived from the distance formula: \[ 5x + 5(x + 15) = 275 \] Here, 5x is the distance traveled by the slower car, and 5(x + 15) is the distance by the faster car. By combining like terms, we get: \[ 10x + 75 = 275 \] Simplify further by isolating x: \[ 10x = 200 \] \[ x = 20 \] Now we've solved for x, which gives us the speed of the slower car. Understanding how to set up and solve equations allows us to determine unknowns in similar problems effectively.