solve for x.\n\n\\( \\frac { 2 } { x - 2 } = \\frac { x } { x - 2 } - \\frac { x } { 3 } \\)\n\nif there are…

solve for x.\n\n\\( \\frac { 2 } { x - 2 } = \\frac { x } { x - 2 } - \\frac { x } { 3 } \\)\n\nif there are two solutions, write one solution in each box. if there is one solution, leave the second box empty.\n\n\\( x = \\square \\) or \\( x = \\square \\)

solve for x.\n\n\\( \\frac { 2 } { x - 2 } = \\frac { x } { x - 2 } - \\frac { x } { 3 } \\)\n\nif there are two solutions, write one solution in each box. if there is one solution, leave the second box empty.\n\n\\( x = \\square \\) or \\( x = \\square \\)

Answer

Explanation:

Step1: Eliminate the denominator

Multiply both sides of the equation $\frac{2}{x - 2}=\frac{x}{x - 2}-\frac{x}{3}$ by $3(x - 2)$ (since $x\neq2$). $$3(x - 2)\times\frac{2}{x - 2}=3(x - 2)\times\frac{x}{x - 2}-3(x - 2)\times\frac{x}{3}$$ Simplify to get $6 = 3x-(x(x - 2))$.

Step2: Expand and rearrange the equation

Expand $-(x(x - 2))$ to $-x^{2}+2x$. The equation becomes $6 = 3x - x^{2}+2x$. Rearrange it to the standard quadratic form $x^{2}-5x + 6=0$.

Step3: Factor the quadratic equation

Factor $x^{2}-5x + 6$. We look for two numbers that multiply to $6$ and add up to $-5$. These numbers are $-2$ and $-3$. So, $x^{2}-5x + 6=(x - 2)(x - 3)=0$. But we note that $x\neq2$ (because the original denominator $x - 2\neq0$).

Answer:

$x = 3$ or $x = $