Question

Determine whether the values are solutions to the equation: $9y+2=6y+3$.

(a) $y=\frac{4}{3}$(b) $y=\frac{1}{3}$.

Short answer

Part (a) $y=\frac{4}{3}$is not a solution to the equation.

Part (b) $y=\frac{1}{3}$is a solution to the equation.

Step-by-step solution

Part (a) Step 1. Given information.

Determine whether the values are solutions to the equation: $9y+2=6y+3$.

(a)$y=\frac{4}{3}$(b)$y=\frac{1}{3}$

Part (a) Step 2. Substitute the value y=43in the equation.

$$\begin{gathered} 9y+2=6y+3 \\ 9\left(\frac{4}{3}\right)+2=6\left(\frac{4}{3}\right)+3 \\ 3·4+2=2·4+3 \\ 14\ne 11 \end{gathered}$$

Thus $y=\frac{4}{3}$does not result in a true equation.

Hence, $y=\frac{4}{3}$is not a solution of the equation.

Part (b) Step 3. Substitute the value y=13in the equation.

$$\begin{gathered} 9y+2=6y+3 \\ 9\left(\frac{1}{3}\right)+2=6\left(\frac{1}{3}\right)+3 \\ 3·1+2=2·1+3 \\ 5=5 \end{gathered}$$

Thus $y=\frac{1}{3}$result in a true equation.

Hence, $y=\frac{1}{3}$is a solution of the equation.

Step 4. Conclusion.

Hence, (a) $y=\frac{4}{3}$is not a solution of the equation and (b) $y=\frac{1}{3}$ is a solution of the equation.