Q 278 - Intermediate Algebra

Question

In the following exercises, solve:

$(5b + 1)(6b + 1) = 0$

Step-by-Step Solution

Answer

The solutions are $b=\frac{-1}{5} and b=\frac{-1}{6}$.

Step 1. Given Information

The given expression is$(5b+1)(6b+1)= 0$.

We have to solve the expression by using the Zero Product Property. 

Step 2. Use the Zero Product Property

If the product equals zero then at least one factor must equal zero.

So, we will put each factor equal to zero.

$$\begin{gathered} (5b+1)(6b+1) = 0 \\ 5b+1=0 or 6b+1=0 \end{gathered}$$

Step 3. Solve the linear equations

The equation we get are

$$\begin{gathered} 5b+1=0 \\ 6b+1=0 \end{gathered}$$

Therefore,

$$\begin{gathered} 5b+1=0 \\ 5b=-1 \\ b=\frac{-1}{5} \end{gathered}$$ and $$\begin{gathered} 6b+1=0 \\ 6b=-1 \\ b=\frac{-1}{6} \end{gathered}$$

Step 4. Verify the answers

To verify the answer we will substitute each solution separately into the original equation.

If $b=\frac{-1}{5}$

$$\begin{gathered} \left[5\right(\frac{-1}{5})+1]\left[6\right(\frac{-1}{5})+1]=0 \\ [-1+1][-\frac{6}{5}+1]=0 \\ 0=0 \end{gathered}$$

If $b=\frac{-1}{6}$

$$\begin{gathered} \left[5\right(\frac{-1}{6})+1]\left[6\right(\frac{-1}{6})+1]=0 \\ [\frac{-5}{6}+1]\left[0\right]=0 \\ 0=0 \end{gathered}$$

Therefore, solutions are right.