The problem involves solving the equation:
$$
\frac{(2x + 3)^6}{6} = \frac{(3x + 1)^6}{6}
$$
Let's solve it step by step.
---
Step 1: Eliminate the denominators
Since both sides of the equation have the same denominator (6), we can eliminate it by multiplying through by 6:
$$
(2x + 3)^6 = (3x + 1)^6
$$
---
Step 2: Analyze the equation
The equation $(2x + 3)^6 = (3x + 1)^6$ implies that the sixth powers of the two expressions are equal. For two numbers to have the same sixth power, they must either be equal or negatives of each other. Therefore, we have two cases to consider:
1. $2x + 3 = 3x + 1$
2. $2x + 3 = -(3x + 1)$
---
Case 1: $2x + 3 = 3x + 1$
Solve for $x$:
$$
2x + 3 = 3x + 1
$$
Subtract $2x$ from both sides:
$$
3 = x + 1
$$
Subtract 1 from both sides:
$$
x = 2
$$
---
Case 2: $2x + 3 = -(3x + 1)$
Solve for $x$:
$$
2x + 3 = -3x - 1
$$
Add $3x$ to both sides:
$$
2x + 3x + 3 = -1
$$
$$
5x + 3 = -1
$$
Subtract 3 from both sides:
$$
5x = -4
$$
Divide by 5:
$$
x = -\frac{4}{5}
$$
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Step 3: Verify the solutions
We have two potential solutions: $x = 2$ and $x = -\frac{4}{5}$. Let's verify them by substituting back into the original equation.
#### Verify $x = 2$:
Substitute $x = 2$ into $(2x + 3)^6 = (3x + 1)^6$:
$$
2x + 3 = 2(2) + 3 = 4 + 3 = 7
$$
$$
3x + 1 = 3(2) + 1 = 6 + 1 = 7
$$
So, $(2x + 3)^6 = 7^6$ and $(3x + 1)^6 = 7^6$, which is true.
#### Verify $x = -\frac{4}{5}$:
Substitute $x = -\frac{4}{5}$ into $(2x + 3)^6 = (3x + 1)^6$:
$$
2x + 3 = 2\left(-\frac{4}{5}\right) + 3 = -\frac{8}{5} + 3 = -\frac{8}{5} + \frac{15}{5} = \frac{7}{5}
$$
$$
3x + 1 = 3\left(-\frac{4}{5}\right) + 1 = -\frac{12}{5} + 1 = -\frac{12}{5} + \frac{5}{5} = -\frac{7}{5}
$$
So, $(2x + 3)^6 = \left(\frac{7}{5}\right)^6$ and $(3x + 1)^6 = \left(-\frac{7}{5}\right)^6$. Since $\left(-\frac{7}{5}\right)^6 = \left(\frac{7}{5}\right)^6$, this is also true.
Both solutions satisfy the original equation.
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Final Answer:
The solutions to the equation are:
$$
\boxed{2 \text{ and } -\frac{4}{5}}
$$
Parent Tip: Review the logic above to help your child master the concept of solving rational equations.
