the three digit base x numeral 7y3 is twice the three-digit base x numeral 3y7. express x+y as a base 10 numeral

We are given the three-digit base-\(x\) numerals \(7y3_x\) and \(3y7_x\), where \(7y3_x\) is twice \(3y7_x\), and we are tasked with finding \(x + y\) in base 10.

### Step 1: Convert both numerals into base-10 expressions.
1. The numeral \(7y3_x\) in base-\(x\) can be expressed as:
   \[
   7x^2 + yx + 3
   \]

2. The numeral \(3y7_x\) in base-\(x\) can be expressed as:
   \[
   3x^2 + yx + 7
   \]

### Step 2: Set up the equation based on the problem.
We are told that \(7y3_x\) is twice \(3y7_x\), so we can write the equation:
\[
7x^2 + yx + 3 = 2(3x^2 + yx + 7)
\]

### Step 3: Expand and simplify the equation.
Expanding the right-hand side:
\[
7x^2 + yx + 3 = 6x^2 + 2yx + 14
\]

Now, subtract \(6x^2 + 2yx + 14\) from both sides:
\[
7x^2 + yx + 3 - 6x^2 - 2yx - 14 = 0
\]
Simplifying:
\[
x^2 - yx - 11 = 0
\]

### Step 4: Solve for \(x\) and \(y\).
We need to find integer values of \(x\) and \(y\) that satisfy the equation \(x^2 - yx - 11 = 0\).

By trial and error, checking values for \(x\):
- For \(x = 5\):
  \[
  5^2 - 5y - 11 = 0 \implies 25 - 5y - 11 = 0 \implies 14 = 5y \implies y = \frac{14}{5}
  \]
  (This is not an integer, so \(x = 5\) doesn't work.)

- For \(x = 6\):
  \[
  6^2 - 6y - 11 = 0 \implies 36 - 6y - 11 = 0 \implies 25 = 6y \implies y = \frac{25}{6}
  \]
  (This is not an integer either.)

- For \(x = 7\):
  \[
  7^2 - 7y - 11 = 0 \implies 49 - 7y - 11 = 0 \implies 38 = 7y \implies y = \frac{38}{7}
  \]



The three-digit base x numeral 7y3 is twice the three-digit base x numeral 3y7. Express x+y as a base 10 numeral.

Learn how to solve the problem where the three-digit base-x numeral 7y3_x is twice the numeral 3y7_x. This solution covers concepts of base conversion, algebraic expressions, and systems of equations, providing a detailed step-by-step explanation. Suitable for Grades 9-12.

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