Question Video: Solving a Word-Problem Using Linear Equations | Nagwa Question Video: Solving a Word-Problem Using Linear Equations | Nagwa

# Question Video: Solving a Word-Problem Using Linear Equations Mathematics • First Year of Preparatory School

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The price of 1 kg of oranges is 2 pounds more than the price of 1 kg of apples. If the price of 2 kg of oranges and 3 kg of apples equals 20 pounds, find the price of 1 kg of each kind.

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### Video Transcript

The price of one kilogram of oranges is two pounds more than the price of one kilogram of apples. If the price of two kilograms of oranges and three kilograms of apples equals 20 pounds, find the price of one kilogram of each kind.

In this question, we are given some information about the costs of various amounts of oranges and apples and asked to use this information to determine the price of one kilogram of apples and one kilogram of oranges. To find the prices, letβs start by saying that one kilogram of apples costs π pounds and one kilogram of oranges costs π pounds. We can then use these unknowns and the given information to construct equations involving π and π. For instance, we are told that one kilogram of oranges is two pounds more expensive than one kilogram of apples. This means that if we add two pounds onto π, the price of one kilogram of apples, then this will be equal to the price of one kilogram of oranges. We have π is equal to π plus two.

In the same way, we can note that two kilograms of oranges will cost two π pounds and three kilograms of apples will cost three π pounds. We are told in the question that buying both of these quantities of fruits costs 20 pounds, so we have three π plus two π is equal to 20. We can construct an equation in terms of only π by substituting our expression for π into the other equation. Substituting π equals π plus two into the equation gives us three π plus two times π plus two is equal to 20.

To solve this equation for π, we want to isolate π on one side of the equation. We can start by distributing the factor of two over the parentheses. We do this by multiplying each term inside the parentheses by two. We obtain three π plus two π plus four is equal to 20. We can combine the like terms on the left-hand side of the equation by noting that three π plus two π equals five π. And we can also subtract four from both sides of the equation. This gives us that five π is equal to 16.

We can now solve for π by dividing both sides of the equation through by five. This gives us that π is 16 over five. And remember, π is the price of one kilogram of apples in pounds. So we know that one kilogram of apples costs 16 over five pounds, which is three pounds 20 pence. We can now find the value of π by substituting π equals 16 over five into our equation for π. This gives us that π is equal to 16 over five plus two. We can then find the value of π by rewriting two as 10 over five to see that π is equal to 26 over five, which means a kilogram of oranges costs five pounds 20 pence.

Hence, we have shown that one kilogram of oranges is 26 over five pounds and one kilogram of apples is 16 over five pounds.

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