Linda left home and drove for 2 hours. She stopped for lunch then .pdf

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: Linda left home and drove for 2 hours. She stopped for lunch then drove for another 3 hours at a rate that is 10 mph higher than the rate before she had lunch. If the total distance Linda travelled is 230 miles, what was the rate before lunch? Solution If x is the rate at which Linda drove before lunch the rate after lunch is equal x + 10. The total distance D traveled by Linda is given by D = 2 x + 3(x + 10) and is equal to 230 miles. Hence 2 x + 3 (x + 10) = 230 Solve for x to obtain x = 40 miles / hour..

: Linda left home and drove for 2 hours. She stopped for lunch then drove for another 3 hours at
a rate that is 10 mph higher than the rate before she had lunch. If the total distance Linda
travelled is 230 miles, what was the rate before lunch?
Solution
If x is the rate at which Linda drove before lunch the rate after lunch is equal x +
10. The total distance D traveled by Linda is given by D = 2 x + 3(x + 10) and is equal to 230
miles. Hence 2 x + 3 (x + 10) = 230 Solve for x to obtain x = 40 miles / hour.

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