Question

Solve the following equation \[0{\text{ }} = {\text{ }}16 + {\text{ }}4\left( {m - 6} \right)\]

Answer 1

Hint:
Polynomials in one variable are algebraic Expressions that consist of terms in the form $a{x^n}$, where n is a non-negative (i.e. positive or zero) integer and a is a real number and called the coefficient of the term. The degree of a polynomial in one of the variables is the largest exponent polynomial.
Firstly simplify Each side is needed we add or subtraction properties to move the variable team to one side and all other terms to the other side. Use multiplication or division This is the quickest and easiest way to approach linear Regulation.

Complete step by step solution:
step-1
Given equation is \[ - {\text{ }}0{\text{ }} = {\text{ }}16 + {\text{ }}4\left( {m - 6} \right)\]
In this Equation, the variable is the letter m. To solve this Equation 9 needs to get them by itself, that is, we need to get m on one side of the equal sign and some number on the other side.

Step-2
Since, we want the first m on the one side this means that we don't o like the plus 16 that is currently on the same side as the m. We need to subtract this 16 to get mid of it. That is a will need to subtract a 16 from both sides.
\[0 - {\text{ }}16{\text{ }} = {\text{ }}16{\text{ }} + {\text{ }}4{\text{ }}\left( {{\text{ }}m{\text{ }}-{\text{ }}6} \right){\text{ }}-{\text{ }}16\]
Here on the left-hand side, 16 is subtracted from zero so we get to and right-hand side 16 subtract of 16 so remaining form of right-hand side \[4\left( {m - 6} \right)\]
\[ - 16 = {\text{ }}4\left( {m - 6} \right)\;\]

step-3
Now, on the right-hand side, u is multiple of (m-6) term. So first we get mid of it. for this, the equation is divided by 4 in both sides
$ - 16/4 = 4/4(m - 6)$
Now to find the value of m. we add 6 on both sides of the Equation.
\[ - 4 + 6 = m - 6 + 6\]
The left side of minus u plus 6 then get plus and on the right-hand side we get only variable m, \[2{\text{ }} = {\text{ }}m\]

Hence, the value of m is\[\;2\]

Note:
You may be instructed to check your solutions at least in the early stages of learning how to solve Equations, To do this checking you need only plug your answer into the original equation and make sure that you come up with a true statement.