Q. 300 - Intermediate Algebra

Question

In the following exercises, solve:

$\left(2y-3\right)\left(3y-1\right)=8y$

Step-by-Step Solution

Answer

On solving the equation, we get $y=\frac{1}{6},3$.

Step 1. Given information.

Consider the given equation,

$\left(2y-3\right)\left(3y-1\right)=8y$

Write the quadratic equation in standard form,

$$\begin{gathered} 6y^2-2y-9y-8y+3=0 \\ 6y^2-19y+3=0 \end{gathered}$$

Step 2. Factor the quadratic expression.

On factoring, we get,

$$\begin{gathered} 6y^2-18y-y+3=0 \\ 6y\left(y-3\right)-\left(y-3\right)=0 \\ \left(6y-1\right)\left(y-3\right)=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} 6y-1=0 \\ y=\frac{1}{6} \end{gathered}$$

Then,

$$\begin{gathered} y-3=0 \\ y=3 \end{gathered}$$

Step 3. Check the answers.

Substitute $y=\frac{1}{6}$ in the given equation,

$$\begin{gathered} \left(2\left(\frac{1}{6}\right)-3\right)\left(3\left(\frac{1}{6}\right)-1\right)=8\left(\frac{1}{6}\right) \\ \left(\frac{1}{3}-3\right)\left(\frac{1}{2}-1\right)=\frac{4}{3} \\ \left(\frac{-8}{3}\right)\left(\frac{-1}{2}\right)=\frac{4}{3} \\ \frac{4}{3}=\frac{4}{3} \end{gathered}$$

This is true.

Substitute $y=3$ in the given equation,

$$\begin{gathered} \left(2\left(3\right)-3\right)\left(3\left(3\right)-1\right)=8\left(3\right) \\ \left(6-3\right)\left(9-1\right)=8\left(3\right) \\ \left(3\right)\left(8\right)=8\left(3\right) \\ 24=24 \end{gathered}$$

This is also true.

Hence, the answer is $y=\frac{1}{6},3$.