Question

solve $\frac{1}{2} + \frac{1}{2x} = \frac{x^2 - 7x + 10}{4x}$ by rewriting the equation as a proportion. which proportion is equivalent to the original equation? $\frac{x + 2}{2x} = \frac{x^2 - 7x + 10}{4x}$ $\frac{x - 1}{2x} = \frac{x^2 - 7x + 10}{4x}$ $\frac{x + 1}{2x} = \frac{x^2 - 7x + 10}{4x}$ name the true solution(s) to the equation. name the extraneous solution(s) to the equation.

Explanation:

Step1: Simplify left-hand side

Combine the fractions on the left:
$$\frac{1}{2} + \frac{1}{2x} = \frac{x}{2x} + \frac{1}{2x} = \frac{x+1}{2x}$$
This confirms the equivalent proportion is $\frac{x+1}{2x}=\frac{x^2 -7x+10}{4x}$.

Step2: Cross-multiply to eliminate denominators

Multiply both sides by $4x$ (note $x
eq0$):
$$4x \cdot \frac{x+1}{2x} = x^2 -7x+10$$
$$2(x+1) = x^2 -7x+10$$

Step3: Expand and rearrange to quadratic form

$$2x + 2 = x^2 -7x +10$$
$$x^2 -9x +8 = 0$$

Step4: Factor the quadratic equation

$$(x-1)(x-8)=0$$
Solutions from factoring: $x=1$ or $x=8$

Step5: Check for extraneous solutions

Test $x=0$: original equation has denominators $2x$ and $4x$, which are undefined at $x=0$, so $x=0$ is excluded.
Test $x=1$:
Left-hand side: $\frac{1}{2}+\frac{1}{2(1)}=1$
Right-hand side: $\frac{1^2 -7(1)+10}{4(1)}=\frac{4}{4}=1$, valid.
Test $x=8$:
Left-hand side: $\frac{1}{2}+\frac{1}{2(8)}=\frac{4}{8}+\frac{1}{16}=\frac{9}{16}$
Right-hand side: $\frac{8^2 -7(8)+10}{4(8)}=\frac{64-56+10}{32}=\frac{18}{32}=\frac{9}{16}$, valid.
Wait correction: Re-check step 2 cross multiply correctly:
Original equivalent proportion $\frac{x+1}{2x}=\frac{x^2-7x+10}{4x}$
Cancel $x$ (x≠0): $\frac{x+1}{2}=\frac{x^2-7x+10}{4}$
Multiply by 4: $2(x+1)=x^2-7x+10$
$2x+2=x^2-7x+10$
$x^2-9x+8=0$, solutions x=1, x=8. Both are valid. Wait, but wait: original equation denominators are 2, 2x, 4x. x=0 is extraneous because it makes denominators zero, even though it wasn't a solution from the quadratic, it's excluded domain.

Answer:

Equivalent proportion: $\frac{x+1}{2x}=\frac{x^2 -7x+10}{4x}$
True solution(s): $x=1$, $x=8$
Extraneous solution(s): $x=0$