Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on you
solve the equation:
\\(\frac{6}{x + 5} = \frac{4}{x - 1}\\)
what are the solutions?
\\(\bigcirc\\) \\(x = 3\\)
\\(\bigcirc\\) \\(x = 13\\) only
\\(\bigcirc\\) \\(x = -17\\)
\\(\bigcirc\\) \\(x = -5, 1\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{6}{x + 5}=\frac{4}{x - 1}\), we can use the cross - multiplication property of fractions. If \(\frac{a}{b}=\frac{c}{d}\) (where \(b
eq0\) and \(d
eq0\)), then \(a\times d=c\times b\).
Applying this property to our equation, we get \(6(x - 1)=4(x + 5)\).

Step2: Expand both sides of the equation

Using the distributive property \(a(b + c)=ab+ac\), we expand the left - hand side: \(6x-6\) and the right - hand side: \(4x + 20\). So our equation becomes \(6x-6 = 4x+20\).

Step3: Isolate the variable \(x\)

Subtract \(4x\) from both sides of the equation: \(6x-4x-6=4x - 4x+20\), which simplifies to \(2x-6 = 20\).
Then add 6 to both sides: \(2x-6 + 6=20 + 6\), so \(2x=26\).
Divide both sides by 2: \(\frac{2x}{2}=\frac{26}{2}\), which gives \(x = 13\).

We also need to check if \(x = 13\) makes the denominators of the original fractions zero. For the denominator \(x + 5\), when \(x = 13\), \(x+5=13 + 5=18
eq0\). For the denominator \(x - 1\), when \(x = 13\), \(x - 1=13-1 = 12
eq0\). So \(x = 13\) is a valid solution.

Answer:

B. \(x = 13\) only