Question

guided practice
what is the solution of $\frac{x - 2}{2} + \frac{2}{x} = \frac{2x + 1}{2x}$?
a. 1 or -1
b. -0.5 only
c. 3 or 1
d. 3 or -3

Explanation:

Step1: Find common denominator

Common denominator is $2x$. Multiply all terms by $2x$:
$x(x-2) + 4 = 2x + 1$

Step2: Expand and simplify

Expand left side: $x^2 - 2x + 4 = 2x + 1$
Rearrange to quadratic: $x^2 - 4x + 3 = 0$

Step3: Factor quadratic

Factor: $(x-1)(x-3) = 0$
Solutions: $x=1$ or $x=3$? Wait, check original equation for $x=3$:
Left: $\frac{3-2}{2}+\frac{2}{3}=\frac{1}{2}+\frac{2}{3}=\frac{7}{6}$
Right: $\frac{2*3+1}{2*3}=\frac{7}{6}$, valid.
Wait, earlier mistake: original multiplication: $\frac{x-2}{2}*2x=x(x-2)$, $\frac{2}{x}*2x=4$, $\frac{2x+1}{2x}*2x=2x+1$. Correct equation: $x^2-2x+4=2x+1$ → $x^2-4x+3=0$ → roots 1 and 3. But option C is 3 or 1. Wait, check again:
Wait no, wait original problem: is it $\frac{2x+1}{2x}$? Yes. Then $x=1$: left $\frac{1-2}{2}+\frac{2}{1}=-\frac{1}{2}+2=\frac{3}{2}$; right $\frac{2*1+1}{2*1}=\frac{3}{2}$, valid. $x=3$: left $\frac{1}{2}+\frac{2}{3}=\frac{7}{6}$; right $\frac{7}{6}$, valid. So answer is C? Wait no, wait the options: C is 3 or 1. Yes.
Wait I made a mistake earlier in step3. Correct factorization: $x^2-4x+3=(x-1)(x-3)=0$ → $x=1$ or $x=3$. So answer is C.

Step1: Multiply by common denominator $2x$

$x(x-2) + 4 = 2x + 1$

Step2: Expand and rearrange to quadratic

$x^2 - 2x + 4 - 2x -1 = 0$ → $x^2 -4x +3=0$

Step3: Factor quadratic equation

$(x-1)(x-3)=0$ → $x=1$ or $x=3$

Step4: Verify solutions (no extraneous roots)

Both $x=1$ and $x=3$ satisfy original equation.

Answer:

A. 1 or -1