Try multiplying both sides of the equation by $x-2$. Make sure you multiply every term on both sides by $x-2$, not just the fractions.

After multiplying by $x-2$, you should no longer have any denominators. Simplify the right-hand side to get an equation of the form $5x+7 = 3x + c$, then solve for $x$.

In Desmos, enter the left side as y = (5x+7)/(x-2) and the right side as y = 3 + 1/(x-2) in two separate lines.

Look for the point where the two graphs intersect. Tap or click on the intersection to see its coordinates.

The $x$-coordinate of the intersection is the solution to the equation (as long as it is not $2$, which would make the denominator zero). Use that $x$-value as your answer.

The denominator $x-2$ cannot be zero, so $x\neq 2$. Any solution we find must not be $2$.

Multiply both sides of the equation by $x-2$ (which is allowed because $x\neq 2$):

The left side simplifies to $5x+7$. On the right side, distribute $(x-2)$:

Simplify the expression $3(x-2)$ on the right side:

so the equation becomes

which simplifies to

Now you have a linear equation in $x$.

Solve $5x+7 = 3x - 5$:

• Subtract $3x$ from both sides: $2x + 7 = -5$.
• Subtract $7$ from both sides: $2x = -12$.
• Divide both sides by $2$: $x = -6$.

Check that this does not violate the restriction $x\neq 2$ (it does not), and it satisfies the original equation, so the solution is $\boxed{-6}$.