Question

solve for b.
\\(\frac{b}{3b + 14} = \frac{1}{b + 8}\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{b}{3b + 14}=\frac{1}{b + 8}\), we use the cross - multiplication property of proportions. If \(\frac{a}{c}=\frac{d}{e}\), then \(a\times e=c\times d\). So, we get \(b(b + 8)=(3b + 14)\times1\).

Step2: Expand both sides

Expanding the left - hand side: \(b(b + 8)=b^{2}+8b\).
Expanding the right - hand side: \((3b + 14)\times1 = 3b+14\).
So, the equation becomes \(b^{2}+8b=3b + 14\).

Step3: Rearrange the equation to standard quadratic form

Subtract \(3b\) and \(14\) from both sides of the equation \(b^{2}+8b=3b + 14\) to get \(b^{2}+8b-3b - 14 = 0\).
Simplify the left - hand side: \(b^{2}+5b - 14 = 0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(- 14\) and add up to \(5\). The numbers are \(7\) and \(-2\) because \(7\times(-2)=-14\) and \(7+( - 2)=5\).
So, \(b^{2}+5b - 14=(b + 7)(b - 2)=0\).

Step5: Solve for \(b\)

Using the zero - product property, if \((b + 7)(b - 2)=0\), then either \(b+7 = 0\) or \(b - 2 = 0\).

• If \(b+7 = 0\), then \(b=-7\).
• If \(b - 2 = 0\), then \(b = 2\).

We need to check for extraneous solutions by substituting back into the original equation:

For \(b=-7\):
Left - hand side: \(\frac{-7}{3\times(-7)+14}=\frac{-7}{-21 + 14}=\frac{-7}{-7}=1\)
Right - hand side: \(\frac{1}{-7 + 8}=\frac{1}{1}=1\)

For \(b = 2\):
Left - hand side: \(\frac{2}{3\times2+14}=\frac{2}{6 + 14}=\frac{2}{20}=\frac{1}{10}\)
Right - hand side: \(\frac{1}{2 + 8}=\frac{1}{10}\)

Answer:

\(b=-7\) or \(b = 2\)