Question
7x+3=7x+3 3(4x+8)=4(3x+6) 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 and show your work.
Answers
GPT-4o mini
Let's examine the two equations provided and solve them step-by-step.
### Equation 1:
\[ 7x + 3 = 7x + 3 \]
This equation simplifies directly:
1. Subtract \(7x\) from both sides:
\[
7x + 3 - 7x = 7x + 3 - 7x
\]
This simplifies to:
\[
3 = 3
\]
Since this is a true statement for all values of \(x\), it indicates that this equation is an identity, meaning it is true for any value of \(x\). Therefore, every value of \(x\) is a solution, including \(x = 3\) and \(x = -2\).
### Equation 2:
\[ 3(4x + 8) = 4(3x + 6) \]
Now, let's solve this equation step-by-step.
1. Distribute both sides:
\[
12x + 24 = 12x + 24
\]
2. Again, subtract \(12x\) from both sides:
\[
12x + 24 - 12x = 12x + 24 - 12x
\]
This simplifies to:
\[
24 = 24
\]
Similar to the first equation, this is also a true statement, indicating that this equation is also an identity. Therefore, it is true for any value of \(x\), including \(x = 3\) and \(x = -2\).
### Conclusion:
Both equations have been shown to be identities, so they have an infinite number of solutions because they are true for all values of \(x\). Thus, they do not have just one or two solutions; they have all real numbers as solutions.
### Verifying with specific values:
- For \(x = 3\):
**For Equation 1**:
\[
7(3) + 3 = 7(3) + 3
\]
\[
21 + 3 = 21 + 3
\]
\[
24 = 24 \quad \text{(True)}
\]
**For Equation 2**:
\[
3(4(3) + 8) = 4(3(3) + 6)
\]
\[
3(12 + 8) = 4(9 + 6)
\]
\[
3(20) = 4(15)
\]
\[
60 = 60 \quad \text{(True)}
\]
- For \(x = -2\):
**For Equation 1**:
\[
7(-2) + 3 = 7(-2) + 3
\]
\[
-14 + 3 = -14 + 3
\]
\[
-11 = -11 \quad \text{(True)}
\]
**For Equation 2**:
\[
3(4(-2) + 8) = 4(3(-2) + 6)
\]
\[
3(-8 + 8) = 4(-6 + 6)
\]
\[
3(0) = 4(0)
\]
\[
0 = 0 \quad \text{(True)}
\]
Both equations are true for \(x = 3\) and \(x = -2\), confirming that the equations can have more than one solution, in fact, infinitely many solutions.
### Equation 1:
\[ 7x + 3 = 7x + 3 \]
This equation simplifies directly:
1. Subtract \(7x\) from both sides:
\[
7x + 3 - 7x = 7x + 3 - 7x
\]
This simplifies to:
\[
3 = 3
\]
Since this is a true statement for all values of \(x\), it indicates that this equation is an identity, meaning it is true for any value of \(x\). Therefore, every value of \(x\) is a solution, including \(x = 3\) and \(x = -2\).
### Equation 2:
\[ 3(4x + 8) = 4(3x + 6) \]
Now, let's solve this equation step-by-step.
1. Distribute both sides:
\[
12x + 24 = 12x + 24
\]
2. Again, subtract \(12x\) from both sides:
\[
12x + 24 - 12x = 12x + 24 - 12x
\]
This simplifies to:
\[
24 = 24
\]
Similar to the first equation, this is also a true statement, indicating that this equation is also an identity. Therefore, it is true for any value of \(x\), including \(x = 3\) and \(x = -2\).
### Conclusion:
Both equations have been shown to be identities, so they have an infinite number of solutions because they are true for all values of \(x\). Thus, they do not have just one or two solutions; they have all real numbers as solutions.
### Verifying with specific values:
- For \(x = 3\):
**For Equation 1**:
\[
7(3) + 3 = 7(3) + 3
\]
\[
21 + 3 = 21 + 3
\]
\[
24 = 24 \quad \text{(True)}
\]
**For Equation 2**:
\[
3(4(3) + 8) = 4(3(3) + 6)
\]
\[
3(12 + 8) = 4(9 + 6)
\]
\[
3(20) = 4(15)
\]
\[
60 = 60 \quad \text{(True)}
\]
- For \(x = -2\):
**For Equation 1**:
\[
7(-2) + 3 = 7(-2) + 3
\]
\[
-14 + 3 = -14 + 3
\]
\[
-11 = -11 \quad \text{(True)}
\]
**For Equation 2**:
\[
3(4(-2) + 8) = 4(3(-2) + 6)
\]
\[
3(-8 + 8) = 4(-6 + 6)
\]
\[
3(0) = 4(0)
\]
\[
0 = 0 \quad \text{(True)}
\]
Both equations are true for \(x = 3\) and \(x = -2\), confirming that the equations can have more than one solution, in fact, infinitely many solutions.