Question

this is a multi - part item. 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 - 1}{2x}=\frac{x^{2}-7x + 10}{4x}\\) \\(\frac{x + 1}{2x}=\frac{x^{2}-7x + 10}{4x}\\) (checked) \\(\frac{x + 2}{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 the left - hand side

First, we simplify the left - hand side of the equation \(\frac{1}{2}+\frac{1}{2x}\). We find a common denominator, which is \(2x\). So \(\frac{1}{2}=\frac{x}{2x}\), and then \(\frac{1}{2}+\frac{1}{2x}=\frac{x}{2x}+\frac{1}{2x}=\frac{x + 1}{2x}\). So the equation \(\frac{1}{2}+\frac{1}{2x}=\frac{x^{2}-7x + 10}{4x}\) can be rewritten as \(\frac{x + 1}{2x}=\frac{x^{2}-7x + 10}{4x}\) (this is the equivalent proportion we already identified).

Step2: Cross - multiply to solve for \(x\)

Cross - multiplying the proportion \(\frac{x + 1}{2x}=\frac{x^{2}-7x + 10}{4x}\) (note that \(x
eq0\) because it makes the denominators zero), we get \(4x(x + 1)=2x(x^{2}-7x + 10)\).
First, expand both sides:
Left - hand side: \(4x(x + 1)=4x^{2}+4x\)
Right - hand side: \(2x(x^{2}-7x + 10)=2x^{3}-14x^{2}+20x\)
So the equation becomes \(4x^{2}+4x=2x^{3}-14x^{2}+20x\).
Bring all terms to one side: \(2x^{3}-14x^{2}+20x-4x^{2}-4x = 0\)
Simplify: \(2x^{3}-18x^{2}+16x = 0\)
Factor out a \(2x\): \(2x(x^{2}-9x + 8)=0\)
Factor the quadratic: \(2x(x - 1)(x - 8)=0\)
Set each factor equal to zero:

• \(2x=0\Rightarrow x = 0\)
• \(x - 1=0\Rightarrow x = 1\)
• \(x - 8=0\Rightarrow x = 8\)

Step3: Check for extraneous solutions

We need to check if these solutions make the original equation's denominators zero. The original equation has denominators \(2\), \(2x\), and \(4x\). When \(x = 0\), the denominators \(2x\) and \(4x\) are zero, so \(x = 0\) is extraneous.
Check \(x = 1\):
Left - hand side: \(\frac{1}{2}+\frac{1}{2(1)}=\frac{1}{2}+\frac{1}{2}=1\)
Right - hand side: \(\frac{1^{2}-7(1)+10}{4(1)}=\frac{1 - 7 + 10}{4}=\frac{4}{4}=1\). So \(x = 1\) is a true solution.
Check \(x = 8\):
Left - hand side: \(\frac{1}{2}+\frac{1}{2(8)}=\frac{1}{2}+\frac{1}{16}=\frac{8 + 1}{16}=\frac{9}{16}\)
Right - hand side: \(\frac{8^{2}-7(8)+10}{4(8)}=\frac{64-56 + 10}{32}=\frac{18}{32}=\frac{9}{16}\). So \(x = 8\) is a true solution.

Answer:

True solution(s): \(x = 1\), \(x = 8\)
Extraneous solution(s): \(x = 0\)