Question

A man says “I am thinking of. When I divide it by \[3\] and then add \[5\], my answer is twice the number I thought of’’’

Answer 1

Hint:Variables (\[{\text{x}},{\text{y,z}}\]etc.) and constant (\[8,5,7\]etc.) combine to make algebraic equation including the operations of addition, subtraction, multiplication and division and equality sign to make the two expression equal
Here the number man is thinking of is variable because it is unknown and hence assume it as \[{\text{x}}\]and further divided by number \[3\](constant) so we get \[\dfrac{{\text{x}}}{3}\] and then \[5\] is added turns into \[\dfrac{{\text{x}}}{3} + 5\] which is equal to the twice of the number (\[{\text{x}}\]) i.e.\[{\text{2x}}\]
Hence we get:
\[\dfrac{{\text{x}}}{3} + 5 = 2{\text{x}}\]
Which is a linear equation


Complete step by step solution:
We are not given the number so we let the number be \[{\text{x}}\].
And algebraic equation formed according to the provided condition is:
\[\dfrac{{\text{x}}}{3} + 5 = 2{\text{x}}\] Algebraic equation to get value of the number, the man was thinking\[\left( {\text{x}} \right)\].
\[ \Rightarrow \dfrac{{\text{x}}}{3} + 5 = 2{\text{x}}\]
Step1
Now solving the above
\[ \Rightarrow \dfrac{{\text{x}}}{3} = 2{\text{x - 5}}\]
\[ \Rightarrow \dfrac{{\text{x}}}{3} - 2{\text{x}} = {\text{ - 5}}\]
\[ \Rightarrow \dfrac{{{\text{x - 6x}}}}{3} = {\text{ - 5}}\]
\[ \Rightarrow \dfrac{{{\text{ - 5x}}}}{3} = {\text{ - 5}}\]
\[ \Rightarrow \dfrac{{{\text{ - 5x}}}}{3} = {\text{ - 5}}\]
\[ \Rightarrow {\text{ x}} = {\text{ - 5}} \times \left( {\dfrac{{ - 3}}{5}} \right)\]
\[ \Rightarrow {\text{ x}} = 3\]
Hence the value of\[{\text{x}}\] is\[3\]
Thus the number of which the man is thinking of is \[3\].


Note: Here there was one variable and was linear (highest power of variable is one) so one equation was enough to solve the problem.
Categories of equation-
(a)There are equations with more than variable.
No. of variables in a problem = no. of equation required to solve the system of equation.
Solved using different methods like – substitution method, elimination method.
(b)There also exist non-linear equation
Which are solved using different methods like- middle term factor, long division method and many others.