We have seen examples of systems of linear equations that have no solutions, one unique solution, and infinitely many solutions.
1 answer:
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Refer to the diagram shown below.
Let the distance that Sassoon travels is d miles.
Let his average speed be x mph.
The time of travel is from 1 p.m. to 4 p.m., which is 3 hours.
Therefore
3x = d (1)
The distance that Archer travels is (d + 300) miles.
The time of travel is from 9 a.m. to 4 p.m., which is 7 hours.
The average traveling speed is (x + 20) mph, therefore
7(x + 20) = d + 300
That is,
7x + 140 = d + 300
7x = d + 160 (2)
Subtract (1) from (2).
7x - 3x = d + 160 - d
4x = 160
x = 40 mph (Sassoon's average speed)
x+20 = 60 mph (Archer's average speed)
From (1), obtain d = 120 mi
Answer:
Archer's speed is 60 mph.
Sassoon's speed is 40 mph.
Let the distance that Sassoon travels is d miles.
Let his average speed be x mph.
The time of travel is from 1 p.m. to 4 p.m., which is 3 hours.
Therefore
3x = d (1)
The distance that Archer travels is (d + 300) miles.
The time of travel is from 9 a.m. to 4 p.m., which is 7 hours.
The average traveling speed is (x + 20) mph, therefore
7(x + 20) = d + 300
That is,
7x + 140 = d + 300
7x = d + 160 (2)
Subtract (1) from (2).
7x - 3x = d + 160 - d
4x = 160
x = 40 mph (Sassoon's average speed)
x+20 = 60 mph (Archer's average speed)
From (1), obtain d = 120 mi
Answer:
Archer's speed is 60 mph.
Sassoon's speed is 40 mph.