Question
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\( 25 a + 10 c - 5 = 25 ( \square \square ) + 10 ( \square \square ) - 5 \)

Ask by Turner Joseph. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The values that replace the squares are \(a\) and \(c\), so the equation becomes: \[ 25a + 10c - 5 = 25a + 10c - 5 \]

Solution

To solve the equation \[ 25 a + 10 c - 5 = 25 ( \square \square ) + 10 ( \square \square ) - 5, \] we can start by simplifying both sides. 1. **Identify the structure of the equation**: The left side is \(25a + 10c - 5\) and the right side has a similar structure with placeholders. 2. **Set the equation without the constant**: We can ignore the \(-5\) on both sides for now, leading to: \[ 25a + 10c = 25 ( \square \square ) + 10 ( \square \square ). \] 3. **Factor out common terms**: We can factor out \(25\) and \(10\) from both sides: \[ 25a + 10c = 25x + 10y, \] where \(x\) and \(y\) are the values that will replace the squares. 4. **Equate coefficients**: This gives us two equations: \[ 25a = 25x \quad \text{and} \quad 10c = 10y. \] 5. **Solve for \(x\) and \(y\)**: From \(25a = 25x\), we can divide both sides by \(25\): \[ a = x. \] From \(10c = 10y\), we can divide both sides by \(10\): \[ c = y. \] Thus, we can conclude that: \[ \boxed{(a, c)}. \] This means that the values that replace the squares are \(a\) and \(c\) respectively.

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Simplify this solution

Mind Expander

To solve this equation, we can simplify both sides. The left side can be rewritten as \( 25(a) + 10(c) - 5 \). If we want to match the right side, we can deduce that \( a \) and \( c \) should be placed in the squares accordingly. So the right side becomes \( 25(a) + 10(c) - 5 \) just with different notation. It's a great reminder that sometimes simplification and matching terms can help in analyzing equations! Always ensure that both sides of the equation are balanced and understood correctly. If you’re diving deeper into algebra, don't forget to explore various identities and properties of equality that can help with similar equations! You can find plenty of engaging resources online or through math clubs that make learning fun and interactive.

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