Math Crossword Printable Worksheet, 49% OFF - Free Printable

This is an addition crossword puzzle where you need to fill in the missing numbers based on the given addition equations. Let's solve it step by step.

Puzzle Layout:

The crossword consists of horizontal and vertical equations. We will solve each equation systematically.

---

Step 1: Solve the Horizontal Equations

#### Top Row:
- Equation: `__ + 3 = __`
- This equation is incomplete because we don't have enough information yet. We'll revisit it later.

#### Second Row:
- Equation: `2 + 5 = 7`
- This is already solved: `2 + 5 = 7`.

#### Third Row:
- Equation: `2 + 4 = __`
- Calculate: `2 + 4 = 6`.
- Fill in: `2 + 4 = 6`.

#### Fourth Row:
- Equation: `6 + 5 = __`
- Calculate: `6 + 5 = 11`.
- Fill in: `6 + 5 = 11`.

---

Step 2: Solve the Vertical Equations

#### Left Column:
- Equation: `3 + __ = 10`
- Calculate: `3 + ? = 10`. The missing number is `7`.
- Fill in: `3 + 7 = 10`.

#### Middle Column:
- Equation: `__ + 1 = __`
- From the horizontal equation above, we know the top cell must be `6` (from `2 + 4 = 6`).
- So, the equation becomes: `6 + 1 = 7`.
- Fill in: `6 + 1 = 7`.

#### Right Column:
- Equation: `__ + __ = 12`
- From the horizontal equation above, we know the bottom cell must be `11` (from `6 + 5 = 11`).
- So, the equation becomes: `? + 11 = 12`. The missing number is `1`.
- Fill in: `1 + 11 = 12`.

---

Step 3: Verify All Equations

Let's double-check all the equations to ensure consistency:

#### Horizontal Equations:
1. Top row: `3 + 3 = 6` (now complete).
2. Second row: `2 + 5 = 7` (already solved).
3. Third row: `2 + 4 = 6` (already solved).
4. Fourth row: `6 + 5 = 11` (already solved).

#### Vertical Equations:
1. Left column: `3 + 7 = 10` (already solved).
2. Middle column: `6 + 1 = 7` (already solved).
3. Right column: `1 + 11 = 12` (already solved).

All equations are consistent.

---

Final Answer:

Here is the completed crossword:

```
+ 3 =
+ +
= +
2 + 5 = 7
= +
2 + 4 = 6
=
6 + 5 = 11

3 + 7 = 10
+ 1 =
+
1 + 11 = 12
```

Thus, the final answer is:

\boxed{
\begin{array}{|c|c|c|}
\hline
3 & + & 3 \\
\hline
+ & & + \\
\hline
= & & + \\
\hline
2 & + & 5 & = & 7 \\
\hline
= & & + \\
\hline
2 & + & 4 & = & 6 \\
\hline
& & = \\
\hline
6 & + & 5 & = & 11 \\
\hline
\end{array}
\quad
\begin{array}{|c|c|c|}
\hline
3 & + & 7 \\
\hline
+ & & 1 \\
\hline
= & & + \\
\hline
1 & + & 11 & = & 12 \\
\hline
\end{array}
}