Question

1. solve and check for extraneous solutions:

show complete work on your paper!
\\(\frac{4}{x} + 1 = \frac{5}{3}\\)
solution: \\(x = \square\\)
\\(\bigcirc\\) this solution is not extraneous.
\\(\bigcirc\\) this solution is extraneous.

Explanation:

Step1: Isolate the rational term

Subtract 1 from both sides.
$\frac{4}{x} = \frac{5}{3} - 1$

Step2: Simplify the right-hand side

Calculate the numerical subtraction.
$\frac{4}{x} = \frac{5}{3} - \frac{3}{3} = \frac{2}{3}$

Step3: Cross-multiply to solve for x

Set product of extremes equal to means.
$2x = 4 \times 3$

Step4: Calculate and solve for x

Divide both sides by 2.
$x = \frac{12}{2} = 6$

Step5: Check for extraneous solutions

Substitute $x=6$ back into original equation.
$\frac{4}{6} + 1 = \frac{2}{3} + 1 = \frac{2}{3} + \frac{3}{3} = \frac{5}{3}$, which matches the right-hand side.

Answer:

$x=6$
This solution is not extraneous.