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# First Degree Equation Applications

### First Degree Equation Problem 1

### First Degree Equation Problem 2

### First Degree Equation Problem 3

### First Degree Equation Problem 4

- 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

- 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

- 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.

- 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.

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