A race car driver is doing a time trial of a new car on a straight track. The car's distance from the timekeeper is represented by y=293x+50, where x is time in seconds and y is distance in feet from the timekeeper's position. How many feet from the timekeeper is the race car at the beginning? It is ft from the timekeeper.

The distance at which the timekeeper is the race car at the start is 50 feet.

Step-by-step explanation:

You know that the car's distance from the timekeeper is represented by

y=293*x +50

where x is time in seconds and y is distance in feet from the timekeeper's position.

You want to determine how far the timekeeper is from the race car at the start. That is, the distance the timekeeper is from the car when the time is equal to zero.  This indicates that x = 0. Replacing x by that value in the expression of the distance of the car from the timekeeper as a function of time and solving:

y=293*0 +50

you get:

y=50

The distance at which the timekeeper is the race car at the start is 50 feet.

so, converting firstly the mixed fractions to improper fractions and then "adding".