Question

1. solve the equation \\(\frac{x + 3}{x - 3} + \frac{x}{x - 5} = \frac{x + 5}{x - 5}\\) by using the lcd. select the solution(s).

Explanation:

Step1: Identify the LCD

The denominators are \(x - 3\) and \(x - 5\), so the LCD is \((x - 3)(x - 5)\). Multiply each term by the LCD to eliminate the fractions:
\[
(x - 3)(x - 5)\cdot\frac{x + 3}{x - 3}+(x - 3)(x - 5)\cdot\frac{x}{x - 5}=(x - 3)(x - 5)\cdot\frac{x + 5}{x - 5}
\]

Step2: Simplify each term

Simplify the left - hand side and the right - hand side:

• For the first term: \((x - 3)(x - 5)\cdot\frac{x + 3}{x - 3}=(x - 5)(x + 3)=x^{2}+3x-5x - 15=x^{2}-2x - 15\)
• For the second term: \((x - 3)(x - 5)\cdot\frac{x}{x - 5}=(x - 3)x=x^{2}-3x\)
• For the right - hand side: \((x - 3)(x - 5)\cdot\frac{x + 5}{x - 5}=(x - 3)(x + 5)=x^{2}+5x-3x - 15=x^{2}+2x - 15\)

Now our equation becomes:
\(x^{2}-2x - 15+x^{2}-3x=x^{2}+2x - 15\)

Step3: Combine like terms

Combine the left - hand side:
\(2x^{2}-5x - 15=x^{2}+2x - 15\)

Step4: Move all terms to one side

Subtract \(x^{2}+2x - 15\) from both sides:
\(2x^{2}-5x - 15-(x^{2}+2x - 15)=0\)
\(2x^{2}-5x - 15 - x^{2}-2x + 15 = 0\)
\(x^{2}-7x=0\)

Step5: Factor the equation

Factor out an \(x\):
\(x(x - 7)=0\)

Step6: Solve for \(x\)

Set each factor equal to zero:

• If \(x = 0\), check the original equation:

Left - hand side: \(\frac{0 + 3}{0 - 3}+\frac{0}{0 - 5}=\frac{3}{-3}+0=-1\)
Right - hand side: \(\frac{0 + 5}{0 - 5}=\frac{5}{-5}=-1\). So \(x = 0\) is a valid solution.

• If \(x-7=0\), then \(x = 7\)

Check \(x = 7\) in the original equation:
Left - hand side: \(\frac{7 + 3}{7 - 3}+\frac{7}{7 - 5}=\frac{10}{4}+\frac{7}{2}=\frac{5}{2}+\frac{7}{2}=\frac{12}{2}=6\)
Right - hand side: \(\frac{7 + 5}{7 - 5}=\frac{12}{2}=6\). So \(x = 7\) is a valid solution.

We also need to check for excluded values. The original equation has denominators \(x - 3\) and \(x - 5\), so \(x
eq3\) and \(x
eq5\). Since \(0\) and \(7\) are not equal to \(3\) or \(5\), they are valid.

Answer:

The solutions of the equation \(\frac{x + 3}{x - 3}+\frac{x}{x - 5}=\frac{x + 5}{x - 5}\) are \(x = 0\) and \(x = 7\)