Solve the equation. Check for extraneous solutions. $\frac { 10 } { x + 2 } - \frac { 1 } { x } = \frac { 1 } { 5 x }$
A) $x = \frac { 6 } { 23 }$
B) $x = \frac { 1 } { 4 }$
C) $x = \frac { 3 } { 14 }$
D)$x = \frac { 3 } { 11 }$
E) $x = \frac { 2 } { 7 }$

Question

Solve the equation. Check for extraneous solutions. $\frac { 10 } { x + 2 } - \frac { 1 } { x } = \frac { 1 } { 5 x }$
A) $x = \frac { 6 } { 23 }$
B) $x = \frac { 1 } { 4 }$
C) $x = \frac { 3 } { 14 }$
D)$x = \frac { 3 } { 11 }$ 
E) $x = \frac { 2 } { 7 }$

Quizplus · Accepted Answer

Multiplying both sides of the equation by $5x(x+2)$ and simplifying gives us $50x=23x^2+46x+20$, which simplifies to $23x^2+4x-20=0$. We can factor this equation to get $(x-2)(23x+10)=0$, so $x=2$ or $x=-\frac{10}{23}$. However, $x=2$ is an extraneous solution since it makes the denominator of the original equation equal to $0$. Thus, the only valid solution is $x=-\frac{10}{23}$, which corresponds to choice D.

Solve the Equation 10x+2−1x=15x\frac { 10 } { x + 2 } - \frac { 1 } { x } = \frac { 1 } { 5 x }x+210​−x1​=5x1​

Solve the Equation $\frac { 10 } { x + 2 } - \frac { 1 } { x } = \frac { 1 } { 5 x }$