Question

Solve the following equation.
\[\dfrac{1}{2}x-3=5+\dfrac{1}{3}x\]

Answer 1

Hint: In this question, we first need to transpose the variable terms to one side and constant terms to the other side. Then simply the obtained equation by rearranging the terms to get the value of x.

Complete step-by-step answer:
Linear Equation: An equation involving the variable in maximum of order 1 is called a linear equation. Graph of a linear equation is a straight line
Linear equation in one variable is of the form \[ax+b=0\]
Now, the steps required for solving linear equation of one variable are
(1) Obtain the linear equation and do cross multiplication if necessary
(2) Transpose the terms involving the variables on the left hand side and those not involving the variables to the right hand side.
(3) Simplify the two sides to obtain the equation of the form \[ax=b\]
(4) Find the value of x as \[x=\dfrac{b}{a}\]
Now, from the given equation in the question we have
\[\dfrac{1}{2}x-3=5+\dfrac{1}{3}x\]
Let us now transpose the variable terms to one side and constant terms to the other side
\[\Rightarrow \dfrac{1}{2}x-\dfrac{1}{3}x=5+3\]
Now, on further simplification of the terms on both the sides we get,
\[\Rightarrow \dfrac{3-2}{6}x=8\]
Now, this can be further written in the simplified form as
\[\Rightarrow \dfrac{1}{6}x=8\]
Now, let us multiply with 6 on both the sides
\[\Rightarrow x=8\times 6\]
Now, on further simplification we get,
\[\therefore x=48\]

Note: Instead of transposing the terms or either sides we can also simplify it by taking L.C.M on the left hand side and right hand side. Then cancel the common terms out and then rearrange the terms to get the result. Both the methods give the same result.
It is important to note that while transposing the terms there is a chance for considering the incorrect sign or neglecting any of the terms. This changes the result completely.