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**Hint:**To three different solutions of $3x-8y=27$, we have to write the equation in terms of either x or y by moving the terms and coefficient of x (or y) to the RHS. Then, we have to substitute different values of y (or x) and find the corresponding values of x (or y).

**Complete step by step solution:**

We have to find three different solutions of $3x-8y=27$ . Let us take 3x to the RHS.

$\Rightarrow -8y=27-3x$

Now, we have to take the coefficient of y to the RHS.

$\begin{align}

& \Rightarrow y=\dfrac{27-3x}{-8} \\

& \Rightarrow y=\dfrac{-\left( 27-3x \right)}{8} \\

\end{align}$

Let us apply distribution property on the numerator of the RHS.

$\Rightarrow y=\dfrac{-27+3x}{8}...\left( i \right)$

Now, let us take different values of x. Let us consider $x=0$ . We have to substitute this value of x in the above equation.

$\begin{align}

& \Rightarrow y=\dfrac{-27+3\times 0}{8} \\

& \Rightarrow y=\dfrac{-27}{8} \\

\end{align}$

Therefore, one solution is $\left( 0,-\dfrac{27}{8} \right)$ .

Now, let us consider $x=1$ . We have to substitute this value of x in equation (i).

$\begin{align}

& \Rightarrow y=\dfrac{-27+3\times 1}{8} \\

& \Rightarrow y=\dfrac{-27+3}{8} \\

\end{align}$

Let us add the terms on the numerator.

$\Rightarrow y=\dfrac{-24}{8}$

We have to divide -24 by 8.

$\Rightarrow y=-3$

Therefore, another solution is $\left( 1,-3 \right)$ .

Now, we have to consider $x=2$ . We have to substitute this value of x in equation (i).

$\begin{align}

& \Rightarrow y=\dfrac{-27+3\times 2}{8} \\

& \Rightarrow y=\dfrac{-27+6}{8} \\

\end{align}$

Let us add the terms on the numerator.

$\Rightarrow y=\dfrac{-21}{8}$

Therefore, another solution is $\left( 2,-\dfrac{21}{8} \right)$ .

**Therefore, three solutions of $3x-8y=27$ are $\left( 0,-\dfrac{27}{8} \right)$ , $\left( 1,-3 \right)$ and $\left( 2,-\dfrac{21}{8} \right)$.**

**Note:**Students must be thorough with algebraic equations and the rules associated with it. We can also find different solutions of $3x-8y=27$ by solving this equation for y.

Let us take -8y to the RHS.

$\Rightarrow 3x=27+8y$

Now, we have to take the coefficient of x to the RHS.

$\Rightarrow x=\dfrac{27+8y}{3}...\left( ii \right)$

Now, let us take different values of y. Let us consider $y=0$ . We have to substitute this value of y in the above equation.

$\begin{align}

& \Rightarrow x=\dfrac{27+8\times 0}{3} \\

& \Rightarrow x=\dfrac{27+0}{3} \\

& \Rightarrow x=\dfrac{27}{3} \\

\end{align}$

We have to divide 27 by 3.

$\Rightarrow x=9$

Therefore, another solution is $\left( 9,0 \right)$ .

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