Question

Plug in x=3 and x=-2 for both of the original equations to show that the equations can have more than one solution. Solve the equations. Explain your work. equation 2: 3x+3=7x+3. equation 6: 3(4x+8)=4(3x+6) dumb it down by a lot

Answer 1

Sure! Let's solve the two equations one by one, then plug in \( x = 3 \) and \( x = -2 \) to see if they hold true.

### Equation 1: \( 3x + 3 = 7x + 3 \)

1. **Start with the equation**:
\( 3x + 3 = 7x + 3 \)

2. **Subtract \( 3 \) from both sides**:
\( 3x = 7x \)

3. **Subtract \( 3x \) from both sides**:
\( 0 = 4x \)

4. **Divide by \( 4 \)**:
\( x = 0 \)

So, the solution for Equation 1 is \( x = 0 \).

### Equation 2: \( 3(4x + 8) = 4(3x + 6) \)

1. **Start with the equation**:
\( 3(4x + 8) = 4(3x + 6) \)

2. **Distribute on both sides**:
Left side: \( 3 \times 4x + 3 \times 8 = 12x + 24 \)
Right side: \( 4 \times 3x + 4 \times 6 = 12x + 24 \)

3. **Now the equation looks like this**:
\( 12x + 24 = 12x + 24 \)

4. **Since both sides are equal, it means**:
This equation is true for all values of \( x \).

So, Equation 2 has infinitely many solutions.

### Plugging in Values

Now let's check if \( x = 3 \) and \( x = -2 \) work for these equations.

For **Equation 1**:
- If \( x = 3 \):
\( 3(3) + 3 = 9 + 3 = 12\)
\( 7(3) + 3 = 21 + 3 = 24\)
**Not equal, so \( x=3 \) is not a solution.**

- If \( x = -2 \):
\( 3(-2) + 3 = -6 + 3 = -3\)
\( 7(-2) + 3 = -14 + 3 = -11\)
**Not equal, so \( x=-2 \) is not a solution.**

For **Equation 2**:
- If \( x = 3 \):
\( 3(4(3) + 8) = 3(12 + 8) = 3(20) = 60\)
\( 4(3(3) + 6) = 4(9 + 6) = 4(15) = 60\)
**Equal, so \( x=3 \) is a solution.**

- If \( x = -2 \):
\( 3(4(-2) + 8) = 3(-8 + 8) = 3(0) = 0\)
\( 4(3(-2) + 6) = 4(-6 + 6) = 4(0) = 0\)
**Equal, so \( x=-2 \) is also a solution.**

### Summary
- Equation 1 has one solution, \( x = 0 \).
- Equation 2 has infinitely many solutions (it’s true for all \( x \)).
- We checked specific values \( x = 3 \) and \( x = -2 \). Only Equation 2 has those as valid solutions, while Equation 1 does not.

This shows that different equations can have different numbers of solutions.