Question

solve for x.
\\(\frac{3}{x + 13} = \frac{-x}{12}\\)

Explanation:

Step1: Cross - multiply the equation

To solve the equation \(\frac{3}{x + 13}=\frac{-x}{12}\), we use the cross - multiplication property of proportions. If \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\). So, we have \(3\times12=(x + 13)\times(-x)\).
Calculating the left - hand side and expanding the right - hand side: \(36=-x^{2}-13x\).

Step2: Rearrange into standard quadratic form

We want to write the equation in the standard quadratic form \(ax^{2}+bx + c = 0\). So, we add \(x^{2}\) and \(13x\) to both sides of the equation \(36=-x^{2}-13x\).
We get \(x^{2}+13x + 36 = 0\).

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(36\) and add up to \(13\). The numbers are \(9\) and \(4\) since \(9\times4 = 36\) and \(9 + 4=13\).
So, the factored form of \(x^{2}+13x + 36\) is \((x + 9)(x+4)=0\).

Step4: Solve for x

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
For \((x + 9)(x + 4)=0\), we set each factor equal to zero:

• If \(x+9 = 0\), then \(x=-9\).
• If \(x + 4=0\), then \(x=-4\).

We need to check these solutions in the original equation to make sure they do not make the denominator zero.
For \(x=-9\): The denominator \(x + 13=-9 + 13 = 4
eq0\), and the right - hand side denominator is \(12
eq0\). Substituting \(x=-9\) into the original equation: \(\frac{3}{-9 + 13}=\frac{3}{4}\), and \(\frac{-(-9)}{12}=\frac{9}{12}=\frac{3}{4}\). So, \(x=-9\) is a valid solution.
For \(x=-4\): The denominator \(x + 13=-4+13 = 9
eq0\), and the right - hand side denominator is \(12
eq0\). Substituting \(x=-4\) into the original equation: \(\frac{3}{-4 + 13}=\frac{3}{9}=\frac{1}{3}\), and \(\frac{-(-4)}{12}=\frac{4}{12}=\frac{1}{3}\). So, \(x=-4\) is also a valid solution.

Answer:

\(x=-9\) or \(x = - 4\)