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) and dumb it down a lot
Answers
GPT-4o mini
Let's start by looking at the two equations you've provided and solving them step by step.
### Equation 1: \( 3x + 3 = 7x + 3 \)
1. **Plug in \( x = 3 \)**:
- Left side: \( 3(3) + 3 = 9 + 3 = 12 \)
- Right side: \( 7(3) + 3 = 21 + 3 = 24 \)
- They are not equal, so \( x = 3 \) is **not** a solution.
2. **Plug in \( x = -2 \)**:
- Left side: \( 3(-2) + 3 = -6 + 3 = -3 \)
- Right side: \( 7(-2) + 3 = -14 + 3 = -11 \)
- They are not equal, so \( x = -2 \) is **not** a solution.
3. **Now let's solve the equation**:
- Start with the original equation: \( 3x + 3 = 7x + 3 \)
- Subtract \( 3 \) from both sides:
\[
3x = 7x
\]
- Subtract \( 3x \) from both sides:
\[
0 = 4x
\]
- Dividing both sides by \( 4 \):
\[
x = 0
\]
### Equation 2: \( 3(4x + 8) = 4(3x + 6) \)
1. **Plug in \( x = 3 \)**:
- Left side: \( 3(4(3) + 8) = 3(12 + 8) = 3(20) = 60 \)
- Right side: \( 4(3(3) + 6) = 4(9 + 6) = 4(15) = 60 \)
- They are equal, so \( x = 3 \) **is a solution**.
2. **Plug in \( x = -2 \)**:
- Left side: \( 3(4(-2) + 8) = 3(-8 + 8) = 3(0) = 0 \)
- Right side: \( 4(3(-2) + 6) = 4(-6 + 6) = 4(0) = 0 \)
- They are equal, so \( x = -2 \) **is a solution**.
3. **Now let's solve the equation**:
- Start with the original equation: \( 3(4x + 8) = 4(3x + 6) \)
- Distributing on both sides:
\[
12x + 24 = 12x + 24
\]
- This shows that both sides are equal for any value of \( x \). This means every \( x \) is a solution.
### Summary:
- For **Equation 1** \( 3x + 3 = 7x + 3 \), the solution is \( x = 0 \).
- For **Equation 2** \( 3(4x + 8) = 4(3x + 6) \), every \( x \) is a solution, so it has infinitely many solutions.
If we test different values like \( x = 3 \) and \( x = -2 \) in **Equation 2**, we'll see they can work, but for **Equation 1** only \( x = 0 \) is a real solution. Thus, we can have different scenarios: one equation may have one solution while another can have infinitely many solutions.
### Equation 1: \( 3x + 3 = 7x + 3 \)
1. **Plug in \( x = 3 \)**:
- Left side: \( 3(3) + 3 = 9 + 3 = 12 \)
- Right side: \( 7(3) + 3 = 21 + 3 = 24 \)
- They are not equal, so \( x = 3 \) is **not** a solution.
2. **Plug in \( x = -2 \)**:
- Left side: \( 3(-2) + 3 = -6 + 3 = -3 \)
- Right side: \( 7(-2) + 3 = -14 + 3 = -11 \)
- They are not equal, so \( x = -2 \) is **not** a solution.
3. **Now let's solve the equation**:
- Start with the original equation: \( 3x + 3 = 7x + 3 \)
- Subtract \( 3 \) from both sides:
\[
3x = 7x
\]
- Subtract \( 3x \) from both sides:
\[
0 = 4x
\]
- Dividing both sides by \( 4 \):
\[
x = 0
\]
### Equation 2: \( 3(4x + 8) = 4(3x + 6) \)
1. **Plug in \( x = 3 \)**:
- Left side: \( 3(4(3) + 8) = 3(12 + 8) = 3(20) = 60 \)
- Right side: \( 4(3(3) + 6) = 4(9 + 6) = 4(15) = 60 \)
- They are equal, so \( x = 3 \) **is a solution**.
2. **Plug in \( x = -2 \)**:
- Left side: \( 3(4(-2) + 8) = 3(-8 + 8) = 3(0) = 0 \)
- Right side: \( 4(3(-2) + 6) = 4(-6 + 6) = 4(0) = 0 \)
- They are equal, so \( x = -2 \) **is a solution**.
3. **Now let's solve the equation**:
- Start with the original equation: \( 3(4x + 8) = 4(3x + 6) \)
- Distributing on both sides:
\[
12x + 24 = 12x + 24
\]
- This shows that both sides are equal for any value of \( x \). This means every \( x \) is a solution.
### Summary:
- For **Equation 1** \( 3x + 3 = 7x + 3 \), the solution is \( x = 0 \).
- For **Equation 2** \( 3(4x + 8) = 4(3x + 6) \), every \( x \) is a solution, so it has infinitely many solutions.
If we test different values like \( x = 3 \) and \( x = -2 \) in **Equation 2**, we'll see they can work, but for **Equation 1** only \( x = 0 \) is a real solution. Thus, we can have different scenarios: one equation may have one solution while another can have infinitely many solutions.