Math Problem Statement
A two-digit number is such that the sum of its digits is = of the number. When the digits of the number are reversed and the number is subtracted from the original number, the result obtained is 45. By letting the tens digit of the original number be x and the ones digit of the original number be y, form two simultaneous equations in terms of x and y in its simplest form and solve them to find the original number.
Solution
Let the two-digit number be represented as , where is the tens digit and is the ones digit. Based on the problem:
Step 1: Form the equations
- Sum of the digits equals the number: Simplify:
ThuslyLet's correctly frame the problem:
Original Problem Restated
Let the two-digit number be , where:
- is the tens digit, and
- is the ones digit.
1. First Condition
The sum of its digits equals one-third of the number: Simplify this equation: 7x - 2y = 0 \tag{1}
2. Second Condition
When the digits are reversed, the resulting number subtracted from the original number is 45: Simplify: x - y = 5 \tag{2}
Step 2: Solve the Equations
We now solve the simultaneous equations:
From (2):
x = y + 5 \tag{3}
Substitute into (1): [ y = -7
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Math Problem Analysis
Mathematical Concepts
Algebra
Simultaneous Equations
Linear Equations
Formulas
Sum of digits equals one-third of the number: x + y = (1/3)(10x + y)
Difference after reversing digits: (10x + y) - (10y + x) = 45
Theorems
Basic principles of linear equations and simultaneous equations
Suitable Grade Level
Grades 8-10
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