Question

which set of algebra tiles represents the equation $x + 2 = 7$?

Explanation:

Step1: Analyze the left side of the equation \( x + 2 = 7 \)

The left side has one \( x \)-tile (representing \( x \)) and two \( 1 \)-tiles (representing \( +2 \)). So we need to find the option where the left side has 1 \( x \)-tile and 2 \( 1 \)-tiles.

Step2: Analyze the right side of the equation \( x + 2 = 7 \)

The right side has seven \( 1 \)-tiles (since \( 7 \) is represented by seven \( 1 \)-tiles). Let's check each option:

• First option: Left side: 1 \( x \)-tile, 1 \( 1 \)-tile (wrong, needs 2 \( 1 \)-tiles). Right side: Let's count the \( 1 \)-tiles. Top row: 3, middle row: 2, bottom row: 2. Total: \( 3 + 2 + 2 = 7 \). But left side has 1 \( 1 \)-tile, not 2.
• Second option: Left side: 1 \( x \)-tile, 2 \( 1 \)-tiles (good). Right side: Top row: 2, middle row: 1, bottom row: 1. Total: \( 2 + 1 + 1 = 4 \) (wrong, needs 7).
• Third option: Left side: 1 \( x \)-tile, 1 \( 1 \)-tile (wrong, needs 2 \( 1 \)-tiles). Right side: Top row: 2, middle row: 1, bottom row: 1. Total: \( 2 + 1 + 1 = 4 \) (wrong).
• Fourth option: Left side: 1 \( x \)-tile, 2 \( 1 \)-tiles (good). Right side: Top row: 3, middle row: 2, bottom row: 2. Total: \( 3 + 2 + 2 = 7 \) (good).

Wait, wait, let's re - check the first option's left side. Wait, the first option's left side: the green \( x \)-tile and one \( 1 \)-tile? No, looking at the first figure: left side is \( x \) (green) and one \( 1 \)-tile? Wait, no, maybe I miscounted. Wait the equation is \( x + 2 = 7 \), so left side: \( x \) plus two \( 1 \)-tiles. Let's re - examine each figure:

1. First figure: Left side: \( x \) (green) and 1 yellow \( 1 \)-tile? No, wait the first figure's left side: the green \( x \)-tile and one yellow \( 1 \)-tile? Wait no, maybe the first figure's left side is \( x \) and one \( 1 \)-tile? Wait no, let's look at the fourth figure: left side: \( x \) and two \( 1 \)-tiles (two yellow \( 1 \)-tiles next to \( x \)). Right side: let's count the \( 1 \)-tiles. Top row: 3, middle row: 2, bottom row: 2. \( 3+2 + 2=7 \). The first figure: left side: \( x \) and one \( 1 \)-tile (wrong, needs two). The fourth figure: left side: \( x \) and two \( 1 \)-tiles (correct for left side), right side: 7 \( 1 \)-tiles (correct). Wait, maybe I made a mistake earlier. Let's re - count each figure:

• First option (top - most):
• Left side: 1 \( x \)-tile, 1 \( 1 \)-tile.
• Right side: Number of \( 1 \)-tiles: 3 (top) + 2 (middle) + 2 (bottom) = 7. But left side has 1 \( 1 \)-tile, not 2. So no.
• Fourth option (bottom - most):
• Left side: 1 \( x \)-tile, 2 \( 1 \)-tiles (two yellow \( 1 \)-tiles next to \( x \)).
• Right side: 3 (top) + 2 (middle) + 2 (bottom) = 7 \( 1 \)-tiles. So this matches \( x + 2 = 7 \) (since left side is \( x+2 \) and right side is \( 7 \)).

Answer:

The fourth set of algebra tiles (the bottom - most one with \( x \), two \( 1 \)-tiles on the left and seven \( 1 \)-tiles on the right) represents the equation \( x + 2 = 7 \). If we assume the options are labeled as the first, second, third, fourth from top to bottom, the answer is the fourth option.