The equation shown in the image is:

$\frac{x}{7 - x} = 2x$

Let's solve for $x$:

Step 1: Eliminate the denominator by multiplying both sides by $7 - x$:

$x = 2x(7 - x)$

Step 2: Distribute $2x$ on the right-hand side:

$x = 14x - 2x^2$

Step 3: Rearrange the equation to bring all terms to one side:

$0 = 14x - 2x^2 - x$

$0 = -2x^2 + 13x$

Step 4: Factor out $x$:

$0 = x(-2x + 13)$

Step 5: Set each factor equal to zero:

1. $x = 0$
2. $-2x + 13 = 0$

Step 6: Solve for $x$ in the second equation:

$-2x = -13$

$x = \frac{13}{2}$

Final Solution:

The possible solutions for $x$ are:

$x = 0 \quad \text{or} \quad x = \frac{13}{2}$

However, we need to check if either solution makes the denominator zero, which would make the expression undefined:

• For $x = 0$, the denominator $7 - 0 = 7$, so it is valid.
• For $x = \frac{13}{2}$, the denominator $7 - \frac{13}{2} = \frac{14}{2} - \frac{13}{2} = \frac{1}{2}$, so it is also valid.

Thus, the solutions are:

$x = 0 \quad \text{or} \quad x = \frac{13}{2}$

Would you like further details or have any questions?

Here are 5 related questions to consider:

1. What happens if the denominator becomes zero during solving an equation?
2. Can you explain how to factor quadratic equations more generally?
3. How can you verify the solution of an equation by substitution?
4. What does the term "extraneous solution" mean in algebra?
5. How does multiplying both sides of an equation help in solving it?

Tip: Always check for restrictions in rational equations, as division by zero can make solutions invalid.