The Difference and Division of Two Numbers
In solving mathematical problems, understanding and employing algebraic equations can be extremely beneficial. In this article, we will delve into a specific problem involving the difference and division of two numbers. We'll discuss how to derive and solve linear equations to find the values of the smaller and larger numbers given certain conditions. Additionally, we'll explore the process step-by-step, making the solution process clear and understandable.The Mathematical Problem
Suppose we have two numbers, where the difference between the larger number ((y)) and the smaller number ((x)) is 1365. Additionally, when the larger number is divided by the smaller number, the quotient is 6, and the remainder is 15. Let's denote the smaller number as (x) and the larger number as (y), and we can establish the following equations: [y - x 1365 quad text{(Equation 1)}] [y 6x 15 quad text{(Equation 2)}] We will explain how to solve these equations step-by-step to find the values of (x) and (y).Solving the Equations
To solve these equations, we will first substitute Equation 2 into Equation 1. [(6x 15) - x 1365] Simplify and combine like terms: [6x 15 - x 1365] [5x 15 1365] Subtract 15 from both sides to isolate the term with (x): [5x 1350] Divide by 5 to solve for (x): [x 270] Now, substitute (x 270) back into Equation 2 to find (y): [y 6(270) 15 1620 15 1635] Thus, the smaller number (x) is 270, and the larger number (y) is 1635.Verification
To ensure our solution is correct, we can verify the following conditions: 1. Check the difference between the larger and smaller number: [y - x 1635 - 270 1365] This confirms the difference condition is met. 2. Check the division condition by dividing the larger number by the smaller number: [y div x 1635 div 270 quad text{gives a quotient of 6 and a remainder of 15}] This confirms the division condition is met, as (1635 div 270 6) with a remainder of 15.Additional Examples
Let's explore additional examples to reinforce our understanding of solving similar problems.Example 1:
Let the greater number (x) and the smaller number (y). The difference between the two numbers is 2395, and when the greater number is divided by the smaller number, the quotient is 6 and the remainder is 15. Here, let's solve these equations: [x - y 2395] [x 6y 15] Substitute (x) from the second equation into the first equation: [6y 15 - y 2395] [5y 15 2395] Subtract 15 from both sides: [5y 2380] Divide by 5: [y 476] Substitute (y 476) back into the second equation to find (x): [x 6(476) 15 2871 15 2886] Thus, the smaller number is 476, and the larger number is 2886.Example 2:
Let smaller number (x). The greater number will be (x 2395). When the greater number is divided by the smaller number, the quotient is 6 and the remainder is 15. Here, the equations are: [x 2395 6x 15] Rearrange to solve for (x): [2395 5x 15] Subtract 15 from both sides: [2380 5x] Divide by 5: [x 476] Thus, the smaller number is 476.