Add one term to this equation so that it results in infinitely many solutions: 

4x + 15 – 9 = 3 + 15x + 5 – 2 

To make the given equation have infinitely many solutions, we need to make both sides of the equation identical. Let's start by simplifying both sides:

### Step 1: Simplify the left-hand side (LHS):
\[
4x + 15 - 9
\]

Combine like terms:
\[
4x + 6
\]

### Step 2: Simplify the right-hand side (RHS):
\[
3 + 15x + 5 - 2
\]

Combine like terms:
\[
15x + 6
\]

### Step 3: Equating both sides to make the equation have infinitely many solutions:

To have infinitely many solutions, both sides of the equation must be identical. We want the LHS to be equal to the RHS. 

From the simplified versions, we have:
\[
4x + 6 = 15x + 6
\]

### Step 4: Finding the term to add:

To make both sides identical, the \(4x\) term on the LHS should match the \(15x\) term on the RHS. To do this, we can add \(11x\) to the LHS:

\[
4x + 11x + 6 = 15x + 6
\]

\[
15x + 6 = 15x + 6
\]

Now, both sides are identical, resulting in infinitely many solutions.

### Conclusion:

Adding the term **\(11x\)** to the left-hand side of the original equation will make it have infinitely many solutions.

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What happens if a different term is added to the equation?
2. How can we determine if an equation has no solutions?
3. Can we make an equation have infinitely many solutions by subtracting a term?
4. What are some examples of equations with infinitely many solutions?
5. How do we solve equations with infinitely many variables?

**Tip:** When checking for infinitely many solutions, look to make both sides of the equation identical.

Add one term to this equation so that it results in infinitely many solutions: 4x + 15 – 9 = 3 + 15x + 5 – 2

Discover how to add a term to the equation 4x + 15 – 9 = 3 + 15x + 5 – 2 to make it have infinitely many solutions. This algebraic problem explores key concepts in creating identical linear equations. Suitable for students in Grades 8-10.

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