Question

question
solve for all values of x.
\\(\frac{x - 1}{x + 5} = \frac{2}{x}\\)
answer attempt 1 out of 2
additional solution no solution
x =
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Explanation:

Step1: Cross - multiply the equation

To solve the rational equation \(\frac{x - 1}{x + 5}=\frac{2}{x}\), 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, we get \(x(x - 1)=2(x + 5)\).

Step2: Expand both sides

Expand the left - hand side: \(x(x - 1)=x^{2}-x\).
Expand the right - hand side: \(2(x + 5)=2x + 10\).
So, the equation becomes \(x^{2}-x=2x + 10\).

Step3: Rearrange into standard quadratic form

Subtract \(2x\) and \(10\) from both sides of the equation to get a quadratic equation in standard form \(ax^{2}+bx + c = 0\).
\(x^{2}-x-2x - 10 = 0\), which simplifies to \(x^{2}-3x - 10 = 0\).

Step4: Factor the quadratic equation

Factor the quadratic expression \(x^{2}-3x - 10\). We need two numbers that multiply to \(- 10\) and add up to \(-3\). The numbers are \(-5\) and \(2\) since \((-5)\times2=-10\) and \(-5 + 2=-3\).
So, \(x^{2}-3x - 10=(x - 5)(x+2)=0\).

Step5: Solve for x and check for extraneous solutions

Set each factor equal to zero:

• If \(x - 5=0\), then \(x = 5\).
• If \(x+2=0\), then \(x=-2\).

Now, we need to check if these solutions make the original denominators zero. The original denominators are \(x + 5\) and \(x\).

• For \(x = 5\): \(x+5=5 + 5=10

eq0\) and \(x = 5
eq0\).

• For \(x=-2\): \(x + 5=-2 + 5=3

eq0\) and \(x=-2
eq0\).

Answer:

\(x = 5\) or \(x=-2\)