How do you solve $8-\left( 4y-7 \right)=1-5y$ ?

Hint: One must remember in algebra, only the terms having the same degree of power of the variable can be added or subtracted. Here, we shift all the terms with the same power on the same side of the equation, and then performing the necessary operations, the value of $y$ can be found.

Complete step-by-step answer:
As we know in algebra, only the terms having the same power on the variable can be added or subtracted.
The given expression is $8-\left( 4y-7 \right)=1-5y$
Now, expand the brackets first according to the BODMAS rule.
$\Rightarrow 8-4y+7=1-5y$
Here, the terms $-5y$ and $-4y$ have the same power of $y$ , and hence they can be added or subtracted from each other.
Similarly, the terms $8,7\;$ and $1$ have the same power of $y$ , which is $0$ because any number with a power of $0$ is equal to $1$ .
Hence, these terms can also be written as $8{{y}^{0}},7{{y}^{0}}$ and ${{y}^{0}}$ . As they have the same power, these terms can be added or subtracted from each other.
To simplify the equation, let us start by shifting the same power terms on the same side of the equation
The equation we are given here is,
$\Rightarrow 8-4y+7=1-5y$
Now, subtracting $15\;$ on both the sides of the equation
$\Rightarrow -4y+8+7-15=-5y+1-15$
$\Rightarrow -4y=-5y-14$
Now, adding $5y\;$ on both the sides of the equation
$\Rightarrow -4y+5y=-5y+5y-14$
$\Rightarrow y=-14$
Hence the solution for the given equation is $y=-14$

Note: Another method for subtracting the terms with $y$ is by taking the common factor i.e., variable $y$ common and subtracting the numbers in the bracket. If one wants to check the accuracy of the solution, one can substitute the obtained value in the given equation and check if it satisfies the equation.
Putting $y=-14$ in the equation.
$\Rightarrow 8-4y+7=1-5y$
$\Rightarrow 8-4\left( -14 \right)+7=1-5\left( -14 \right)$
Now evaluate.
$\Rightarrow 8+56+7=1+70$
$71=71$
Since LHS=RHS, our answer is hence correct.