Question

select the correct choice below and, if necessary, fill
○ a. x =
(type an integer or a fraction. use a comma
○ b. there is no solution.

1. solve the rational equation. if the equation has no so

\\(\frac{8}{x + 3} = \frac{5}{x - 9}\\)
select the correct choice below and, if necessary, fil
○ a. x = (use a comma to separ
○ b. there is no solution.

1. solve the equation.

\\(\frac{1}{x - 18} - \frac{3x}{x^2 - 324} = \frac{4}{x + 18}\\)
select the correct choice below and fill in any ans

Explanation:

Problem 7: Solve \(\frac{8}{x + 3}=\frac{5}{x - 9}\)

Step 1: Cross - multiply

To solve the rational equation \(\frac{8}{x + 3}=\frac{5}{x - 9}\), we use the cross - multiplication property of proportions. If \(\frac{a}{b}=\frac{c}{d}\) (where \(b
eq0\) and \(d
eq0\)), then \(a\times d=c\times b\).
So, \(8(x - 9)=5(x + 3)\)

Step 2: Expand both sides

Expand the left - hand side: \(8x-72\)
Expand the right - hand side: \(5x + 15\)
The equation becomes \(8x-72 = 5x+15\)

Step 3: Isolate the variable \(x\)

Subtract \(5x\) from both sides: \(8x-5x-72=5x - 5x+15\), which simplifies to \(3x-72 = 15\)
Add 72 to both sides: \(3x-72 + 72=15 + 72\), so \(3x=87\)

Step 4: Solve for \(x\)

Divide both sides by 3: \(x=\frac{87}{3}=29\)

We need to check if \(x = 29\) makes the denominators zero. For the denominator \(x + 3\), when \(x = 29\), \(x+3=29 + 3=32
eq0\). For the denominator \(x - 9\), when \(x = 29\), \(x - 9=29-9 = 20
eq0\). So \(x = 29\) is a valid solution.

First, factor the denominator \(x^{2}-324\). We know that \(a^{2}-b^{2}=(a + b)(a - b)\), so \(x^{2}-324=x^{2}-18^{2}=(x + 18)(x - 18)\)

Step 1: Find the common denominator

The common denominator of the three fractions \(\frac{1}{x - 18}\), \(\frac{3x}{(x + 18)(x - 18)}\) and \(\frac{4}{x + 18}\) is \((x + 18)(x - 18)\) (where \(x
eq18\) and \(x
eq - 18\))

Step 2: Rewrite each fraction with the common denominator

\(\frac{1\times(x + 18)}{(x - 18)(x + 18)}-\frac{3x}{(x + 18)(x - 18)}=\frac{4\times(x - 18)}{(x + 18)(x - 18)}\)

Step 3: Simplify the numerators

The left - hand side numerator: \((x + 18)-3x=x + 18-3x=-2x + 18\)
The right - hand side numerator: \(4(x - 18)=4x-72\)
The equation becomes \(\frac{-2x + 18}{(x + 18)(x - 18)}=\frac{4x-72}{(x + 18)(x - 18)}\)

Since the denominators are the same (and non - zero for \(x
eq\pm18\)), we can set the numerators equal to each other:
\(-2x + 18=4x-72\)

Step 4: Isolate the variable \(x\)

Add \(2x\) to both sides: \(18=4x+2x-72\), so \(18 = 6x-72\)
Add 72 to both sides: \(18 + 72=6x\), so \(90 = 6x\)

Step 5: Solve for \(x\)

Divide both sides by 6: \(x = 15\)

Check if \(x = 15\) makes the denominators zero. For \(x-18=15 - 18=-3
eq0\), for \(x + 18=15 + 18 = 33
eq0\), and for \(x^{2}-324=15^{2}-324=225-324=-99
eq0\). So \(x = 15\) is a valid solution.

Answer:

A. \(x = 29\)

Problem 8: Solve \(\frac{1}{x - 18}-\frac{3x}{x^{2}-324}=\frac{4}{x + 18}\)