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Write and Solve the Equation for Each Model (Algebra Tiles)

How can a tile-based model be translated into a linear equation, and what is the solution for the variable in each model?

Subject: Math Algebra Chapter: Equations Topic: Linear Equation Solver Answer included
write and solve the equation for each model algebra tiles equation model linear equation one-variable equation balance model inverse operations integer tiles
Accepted answer Answer included

“Write and solve the equation for each model” means translating a visual equality into algebra. Each side of the model represents a sum of tiles; equality means the two sums are the same number. Once the equation is written, the solution is the value of the variable that makes both sides equal.

Assumption used throughout: the model uses three tile types—one variable tile and two integer tiles.

Tile in the model Algebraic meaning Numeric value
Variable tile labeled x One unknown quantity \(x\)
Positive unit tile labeled +1 Adds one to the sum \(+1\)
Negative unit tile labeled −1 Subtracts one from the sum \(-1\)

Each side of the model becomes an expression formed by adding its tiles. The equality sign is the balance point: \[ \text{(sum of left tiles)} = \text{(sum of right tiles)}. \]

Model-to-equation structure

A variable tile contributes \(x\) each time it appears. Positive and negative unit tiles contribute \(+1\) and \(-1\). Multiple tiles of the same type add up as repeated addition: \[ x + x + x = 3x, \qquad (+1)+(+1)+(+1)=3, \qquad (-1)+(-1)=-2. \]

Model A x + 3 = 7 x +1 +1 +1 = +1 +1 +1 +1 +1 +1 +1 Model B 2x − 2 = 4 x x −1 −1 = +1 +1 +1 +1 Model C x − 5 = −1 x −1 −1 −1 −1 −1 = −1
Each row is a separate model. The left side and right side represent equal totals; each tile contributes \(x\), \(+1\), or \(-1\). Writing the equation means summing the tiles on each side exactly as shown.

Equations and solutions for the models

Model Equation written from the tiles Solution for \(x\) Check (substitution)
Model A \(x + 3 = 7\) \(x = 4\) \(4 + 3 = 7\)
Model B \(2x - 2 = 4\) \(x = 3\) \(2\cdot 3 - 2 = 4\)
Model C \(x - 5 = -1\) \(x = 4\) \(4 - 5 = -1\)
Algebraic justification (rigorous equalities)

Model A. \[ x+3=7 \quad\Rightarrow\quad x=7-3 \quad\Rightarrow\quad x=4. \]

Model B. \[ 2x-2=4 \quad\Rightarrow\quad 2x=4+2 \quad\Rightarrow\quad 2x=6 \quad\Rightarrow\quad x=\frac{6}{2}=3. \]

Model C. \[ x-5=-1 \quad\Rightarrow\quad x=-1+5 \quad\Rightarrow\quad x=4. \]

Sign conventions and common errors

A negative unit tile contributes \(-1\) each time it appears, so five negative tiles add to \(-5\), not \(+5\). Multiple variable tiles add as \(x+x=2x\), not \(x^2\). Equality remains true only when identical operations are applied to both sides, which is the algebraic counterpart of maintaining balance in the model.

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