Solving the Problem of the Difference Between Two Numbers
In this article, we will focus on solving a specific algebraic problem where the difference between two numbers is a given value. This type of problem is common in mathematics and often appears in standardized tests and academic assessments. The problem we will solve is: 'The difference between two numbers is 25. The smaller number is 1/6th of the larger number. What is the value of the smaller number?' Let's break down the problem and solve it step by step.
Setting Up the Equations
The problem can be described using two equations. Let the smaller number be denoted by x and the larger number by y.
The first equation is based on the difference between the two numbers:
x - y 25
The second equation is based on the fact that the smaller number is 1/6th of the larger number:
x y/6
Solving the Equations
We can substitute the second equation into the first equation to solve for y.
x - y 25
Substituting x y/6 into the first equation:
y/6 - y 25
Multiplying through by 6 to clear the fraction:
y - 6y 150
-5y 150
Solving for y:
y -30
Now we can substitute y -30 back into the first equation to find the value of x:
x - (-30) 25
x 30 25
x -5
Therefore, the value of the smaller number is -5.
Another Approach
To ensure the problem is solved accurately, let's consider another method. We can start by assuming the larger number as x and the smaller number as y.
The equations based on the given information are:
y - x 25
y x/6
Substituting the second equation into the first equation:
x/6 - x 25
Multiplying through by 6 to clear the fraction:
x - 6x 150
-5x 150
Solving for x:
x -30
Substituting x -30 into the second equation to find y:
y -30/6
y -5
Thus, reaffirming that the value of the smaller number is -5.
Consideration of Sign
In this context, the equations and solutions show that the smaller number can be negative. In real-world applications, the context of the problem may dictate whether the numbers can be negative. For instance, in financial contexts, negative numbers may not be relevant. However, in the scope of algebraic problem-solving, negative solutions are valid and consistent with the given equations.
In conclusion, the value of the smaller number is -5, based on the given conditions and solving the equations accordingly.