A system of linear equations involves two or more equations that describe the relationship between two unknown values. Each equation represents a different condition, and together, they help determine the values of both variables.
Once we set up the system of equations, we can solve it algebraically, either by substitution or elimination method.
In this article, we will explore real-life scenarios that can be modeled using a system of linear equations and learn how to solve them step by step.
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A theater sold 250 tickets for a show. Adult tickets cost \$10 each, while children’s tickets cost \$5 each. The total revenue from ticket sales was \$2000. How many adult and child tickets were sold?
Solution:
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Let x be the number of adult tickets sold.
Let y be the number of child tickets sold.
Total tickets is: x + y = 250 …..(i)
Total revenue is: 10x + 5y = 2000 …..(ii)
Here, we will solve by substitution.
Solving for y from the first equation,
y = 250 – x
Substituting into the second equation, we get
10x + 5(250 – x) = 2000
⇒ 10x + 1250 – 5x = 2000
⇒ 5x = 750
⇒ x = 150
Substituting x = 150 into y = 250 – x, we get
⇒ y = 250 – 150 = 100
Thus, 150 adult tickets and 100 child tickets were sold.
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A chemist needs to prepare 500 mL of a 30% acid solution by mixing a 40% acid solution with a 20% acid solution. How much of each solution should be used?
Solution:
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Here, we will solve by elimination.
Let x be the volume of 40% solution in ml.
Let y be the volume of 20% solution in ml.
Total volume is: x + y = 500 …..(i)
Total acid content is: 0.40x + 0.20y = 0.30(500) …..(ii)
Multiplying the second equation by 10 to eliminate decimals, we get
4x + 2y = 1500
Multiplying the first equation by 2, we get
2x + 2y = 1000
Subtracting,
(4x + 2y) – (2x + 2y) = 1500 – 1000
⇒ 2x = 500
⇒ x = 250
Substituting x = 250 into x + y = 500, we get
⇒ 250 + y = 500
⇒ y = 250
Thus, 250 ml of 40% solution and 250 ml of 20% solution are needed.
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The sum of two numbers is 50. The difference between three times the larger number and twice the smaller number is 60. What are the two numbers?
Solution:
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Here, we will solve by substitution.
Let x be the larger number.
Let y be the smaller number.
The sum of numbers is:
x + y = 50 …..(i)
The difference equation is:
3x – 2y = 60 …..(ii)
Solving for y, we get
⇒ y = 50 – x
Substituting in the second equation, we get
3x – 2(50 – x) = 60
⇒ 3x – 100 + 2x = 60
⇒ 5x = 160
⇒ x = 32
Substituting x = 32 into y = 50 – x, we get
⇒ y = 50 – 32 = 18
Thus, the two numbers are 32 and 18.
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A person invests a total of \$5000 in two different accounts. One account earns 5% annual interest, while the other earns 7% annual interest. At the end of the year, the total interest earned was \$310. How much was invested in each account?
Solution:
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Here, we will solve by substitution.
Let x be the amount invested at 5% interest.
Let y be the amount invested at 7% interest.
Total investment is:
x + y = 5000 …..(i)
Total interest earned is:
0.05x + 0.07y = 310 …..(ii)
Solving for y in terms of x, we get
⇒ y = 5000 – x
Substituting in the second equation, we get
0.05x + 0.07(5000 – x) = 310
⇒ 0.05x + 350 – 0.07x = 310
⇒ -0.02x + 350 = 310
⇒ -0.02x = -40
⇒ x = 2000
Substituting x = 2000 into y = 5000 – x, we get
⇒ y = 5000 – 2000 = 3000
Thus, \$2000 was invested at 5%, and \$3000 was invested at 7%.
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A boat travels 36 km downstream in 3 hours, and the same boat travels 24 km upstream in 4 hours. Find the speed of the boat in still water and the speed of the current.
Solution:
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Here, we will solve by elimination.
Let b be the speed of the boat in still water (km/hr).
Let c be the speed of the current (km/hr).
Downstream speed (speed of the boat + speed of the current) is:
b + c = 36 ÷ 3 = 12 …..(i)
Upstream speed (speed of the boat – speed of the current) is:
b – c = 24 ÷ 4 = 6 …..(ii)
Adding both equations, we get
(b + c) + (b – c) = 12 + 6
⇒ 2b = 18
⇒ b = 9
Substituting b = 9 into b + c = 12, we get
⇒ 9 + c = 12
⇒ c = 3
Thus, the speed of the boat in still water is 9 km/hr, and the speed of the current is 3 km/hr.
