Question

and identify any extraneous roots should they occur.

1. $\frac{x+2}{x-7}=\frac{x}{x-3}$

Explanation:

Step1: State variable restrictions

Denominators cannot be zero, so $x
eq 7$ and $x
eq 3$.

Step2: Cross-multiply to eliminate fractions

$$(x+2)(x-3) = x(x-7)$$

Step3: Expand both sides

Left: $x^2 - 3x + 2x - 6 = x^2 - x - 6$
Right: $x^2 - 7x$
$$x^2 - x - 6 = x^2 - 7x$$

Step4: Simplify to solve for $x$

Subtract $x^2$ from both sides:
$$-x - 6 = -7x$$
Add $7x$ to both sides:
$$6x - 6 = 0$$
$$6x = 6$$
$$x = 1$$

Step5: Check for extraneous roots

$x=1$ is not equal to 7 or 3, so it is valid.

Answer:

Restrictions: $x
eq 3$ and $x
eq 7$
Solution: $x=1$ (no extraneous roots)