How do we solve the following linear system: $8x+2y=3,2x+7=-7y$ ?

Hint: For solving the given linear equations $8x+2y=3$ and $2x+7=-7y$. We will use the substitution method in which first of all we will write one of the equations in terms of $x$ for $y$ in such a way as:
$y=ax+b$ , where $a$ and $b$ are constant. After that we will substitute that variable with $ax+b$ for $y$ in another equation. Then we will solve the equation to get the value of $x$ . After having the value of $x$ , we can use that value in the previous equation to get the value of $y$ .

Complete step by step solution:
The question have two linear equations as:
$\Rightarrow 8x+2y=3$ … $\left( i \right)$
And
$\Rightarrow 2x+7=-7y$ … $\left( ii \right)$
Since, we have two equations we can start with any equation to change the equation in terms of $x$ for $y$ . Here, we will chose equation $\left( ii \right)$ because it is already in terms of $x$ as:
$\Rightarrow -7y=2x+7$
We will divide by $-7$in the above equation. So, the above equation will be as:
$\Rightarrow \dfrac{-7y}{7}=\dfrac{2x+7}{7}$
Now, we will expand the above equation as:
$\Rightarrow \dfrac{-7y}{-7}=\dfrac{2x}{-7}+\dfrac{7}{-7}$
Here, we will do necessary calculation for above equation to change it in terms of $x$ as:
$\Rightarrow y=-\dfrac{2}{7}x-1$ … $\left( iii \right)$
Since, we have equation in terms of $x$, we will use this equation $\left( iii \right)$ in equation $\left( i \right)$ by replacing $y$ to get the value of $x$ as:
$\Rightarrow 8x+2\left( -\dfrac{2}{7}x-1 \right)=3$
Now, we expand the above equation and will do required calculation as:
$\Rightarrow 8x-\dfrac{4}{7}x-2=3$
Here, we can see that $8x$ and $\dfrac{4}{7}x$ are equal like terms. So we combine them where we will subtract $\dfrac{4}{7}x$ from $8x$ and get the value as:
$\Rightarrow \dfrac{7\times 8x-4x}{7}-2=3$
$\Rightarrow \dfrac{56x-4x}{7}-2=3$
$\Rightarrow \dfrac{52x}{7}-2=3$
Since, $2$ and $3$ are numbers, we will place them one side of the equal sign as:
$\Rightarrow \dfrac{52x}{7}=3+2$
We will add $3$ and $2$ :
$\Rightarrow \dfrac{52x}{7}=5$
Here, we will multiply by $7$ in the above equation as:
$\Rightarrow \dfrac{52x}{7}\times 7=5\times 7$
After multiplication, we can get the value as:
$\Rightarrow 52x=35$
Now, we will have divide by $52$ in the above equation as:
$\Rightarrow \dfrac{52x}{52}=\dfrac{35}{52}$
So, we will get the value of $x$ as:
$\Rightarrow x=\dfrac{35}{52}$
Since, we got the value of $x$, we will use this value in equation $\left( iii \right)$ :
$\Rightarrow y=-\dfrac{2}{7}\times \dfrac{35}{52}-1$
Here, we will do necessary calculation for getting the value of $y$from equation as:
$\Rightarrow y=-\dfrac{5}{26}-1$
$\Rightarrow y=\dfrac{-5-1\times 26}{26}$
$\Rightarrow y=\dfrac{-5-26}{26}$
$\Rightarrow y=-\dfrac{31}{26}$
Hence, we have the values $\dfrac{35}{52}$ and $-\dfrac{31}{26}$ for $x$ and $y$ respectively.

Note: Here, we can check or verify the values $\dfrac{35}{52}$ and $-\dfrac{31}{26}$ for $x$ and $y$ respectively in any equation. Let’s check if these values are correct or not by putting this value in the equation $\left( i \right)$. Since the equation is:
$\Rightarrow 8x+2y=3$
Now, we will put the values of $x$ and $y$ in above equation as;
$\Rightarrow 8\times \dfrac{35}{52}+2\left( -\dfrac{31}{26} \right)=3$
Here, we will do necessary calculation as:
$\Rightarrow \dfrac{140}{26}-\dfrac{62}{26}=3$
$\Rightarrow \dfrac{140-62}{26}=3$
$\Rightarrow \dfrac{78}{26}=3$
$\Rightarrow 3=3$
Here, we see that both the values are equal. Hence, the solution is correct.