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Algebra
First Degree Equation Applications
First Degree Equation Problem 1
Question:
A tank was filled with water. One-third of water was removed from the tank. When additional 10 gallons of water added into the tank, there was 20 gallons of water in the tabk. How many of water was in the tank originally?
Solution
let x be the gallons of water in the tank original, after using 1/3 of it
the water in the tank is: x - (1/3)x, then added 10 gallons of water,
the water in the tank is: x - (1/3)x + 10, now left 20 gallons of water in the tank
equation for this problem: x - (1/3)x + 10 = 20
move the number to the right side of the equation
x - (1/3)x = -10 + 20
(3/3)x - (1/3)x = 10
(2/3)x = 10
both side of the equation multiply 3
2x = 10 × 3
2x = 30
both sides of the quation divide by 2
x = 15
therefore, the tank has 15 gallons of water originally
First Degree Equation Problem 2
Question
A and B are two cities that is 100 miles apart. A car from city A moves at the speed of 20 miles per hours and another car from city B moves at the speed of 30 miles per hour toward each other. If two cars start to drive at the same time, how long does it take for the two cars meet together?
Solution
let t be the time that both cars drive (note: distance = time × speed)
then the distance that the car drive from the city A is: 20 t
the distance that the car drive from city B is: 30 t
since bith cars drive facing each other, then the sum of the distance that both cars drived is 100 miles
20 t + 30 t = 100
50 t = 100
divide 50 on both side of the equation
t = 2
therefore, after 2 hours, two cars meet together
First Degree Equation Problem 3
Question
The distance between city A and city B is 100 miles. A car from city A toward city B with speed of 20 miles per hour. 30 minuters later, another car drives from city B toward city A with the speed of 30 miles per hour, when does the two cars meet together?
Solution
let t be the time that the car drive from the city B. Since the car drive from the city A is 30 minutes earlier, then it spend ( t + 30/60)hours
the distance that the car drive from city B is 30 t
the distance that the car drive from city A is 20 (t + 30/60)
the total distance that both cars drive is 100 miles
we get the equation:
20 (t + 30/60) + 30 t = 100
remove parenthesis
20 t + 20 × (30/60) + 30 t = 100
20 t + 10 + 30 t = 100
move the number to the right side of the equation
20 t + 30 t = - 10 + 100
50 t = 90
divide by 50 in both side of the equation
t = 9/5 = 1.8 hours = 1 hour + 48 minutes
therefore, if car from the city A drive first, after 1 hour and 48 minutes, two cars meet together.
First Degree Equation Problem 4
Question
If both cars start at the same time and drive to the same direction. How long does it take when they meet together?
Solution
let t be the time that both cars drive
the distance that the car drive from city A is 20 t
the distance that the car drive from city B is 30 t
since both cars drive at the same direction, then
20 t + 100 = 30 t
move the variables to the left side of the equation and move the numbers to the right side of the equation
20 t - 30 t = - 100
- 10 t = - 100
divide by - 10 on both side of the equation
t = 10
therefore, after 10 hours, the two cars will meet together.