Solve the equation $8s-10=3\left(6-2s\right)$. Also check your solution

Variables: - In any expression or equation, the variable is an unknown quantity. It may be any alphabet or any special symbol.

In $8s-10=3\left(6-2s\right)$, s is the variable which is the unknown quantity.

Constants: - In any expression or equation constants are the values that are fixed.

Also, $8s$and$2s$ are the product of a constant and variable.

While performing addition in an equation, the same kind of elements or like terms are added together.

In the given equation, terms with the variable swill be added all together and while performing addition, constant terms which are multiplied with the variables are added.

The distributive law of multiplication over addition or subtraction is given in below.

$\left(b\pm c\right)a=ba\pm ca$

Here, $a,b$ and$c$ are real numbers or variables respectively.

$8s-10=3\left(6-2s\right)$

Apply distributive law in $3\left(6-2s\right)$and simplify

$\begin{array}{l}8s-10=3\left(6-2s\right)\\ 8s-10=3\left(6\right)-3\left(2s\right)\\ 8s-10=18-6s\end{array}$

Add $6s+10$in both sides of the equation and simplify.

$\begin{array}{l}8s-10+6s+10=18-6s+6s+10\\ 8s+6s-10+10=18+10-6s+6s\\ 8s+6s=18+10\\ 14s=28\end{array}$

Multiply both sides by $\frac{1}{14}$and simplify.

$\begin{array}{l}\frac{1}{14}\times 14s=\frac{1}{14}\times 28\\ s=2\end{array}$

Hence, on solving$8s-10=3\left(6-2s\right)$ the obtained value of s is 2.

Put $s=2$in $8s-10=3\left(6-2s\right)$and check whether the value on the right-hand side is equal to the value on the left-hand side.

$8\left(2\right)-10=3\left(6-2\left(2\right)\right)$

First, evaluate the value on the left-hand side.

$\begin{array}{c}8\left(2\right)-10=16-10\\ =6\end{array}$

Now evaluate the values of right- hand side, that is, $3\left(6-2\left(2\right)\right)$.

$\begin{array}{c}3\left(6-2\left(2\right)\right)=3\left(6-4\right)\\ =3\left(2\right)\\ =6\end{array}$

Since the values of both sides are equal, the obtained result is true and verified.