Question

Solve the following equations:
a)\[3x + 2\left( {x + 2} \right) = 20\left( {2x - 5} \right)\]

Answer 1

Hint: We know that an equation is a statement of equality of two algebraic expressions involving one or more unknown quantities called the variables. A linear equation is an equation involving only a linear polynomial that is the degree of the unknown variable is one.

Complete step-by-step answer:
The equation given above in the question is an example of a linear equation in one variable because it has only one unknown variable ‘$x$’.
Therefore the solution of a linear equation is given by the value of the unknown variable which when substituted for the variable makes both the sides of the given equation equal. This value of the variable is called a solution of the equation. It is also known as the root of the equation.
We can solve the linear equation by its properties of adding, subtracting, multiplying and dividing the quantities on both the sides of the equation such that by these operations the equation remains unchanged. Solving this equation we get,
Firstly let us dissolve the brackets by multiplying and opening the brackets to form resultant terms,
\[3x + 2x + 4 = 40x - 100\]
Adding 100 on both the sides of the equation,
$
\Rightarrow 3x + 2x + 4 + 100 = 40x - 100 + 100 \\
\Rightarrow 5x + 104 = 40x \;
$
Subtracting 5x from both the sides of the equation
$
\Rightarrow 5x + 104 - 5x = 40x - 5x \\
\Rightarrow 104 = 35x \;
$
Dividing by 35 on both the sides of the equation
$
\Rightarrow \dfrac{{104}}{{35}} = \dfrac{{35}}{{35}}x \\
  x = 2.97 \;
$
Hence the value of $x$=2.97 approximately $x$=3.

Note: Further transposition can also be done, it is a process of moving any term of the equation from L.H.S to R.H.S or R.H.S to L.H.S of the equality by changing its sign from positive to negative and negative to positive.