Question

How do you solve $\dfrac{1}{3}x+2=\dfrac{5}{6}$ ?

Answer 1

Hint: We are given a one-degree polynomial equation in x-variable which has one fractional term of x-variable and two constant terms, one each on the right-hand side and the left-hand side. We shall first transpose all the constant terms to the right hand side and simply add or subtract them according to the equation formed. Then we shall multiply the equation with 3 to find the final value of x.

Complete step by step solution:
Given that $\dfrac{1}{3}x+2=\dfrac{5}{6}$
The constant term on the left hand side is 2 and the constant term on the right hand side is $\dfrac{5}{6}$ . We shall transpose 2 to the right hand side first and subtract it from $\dfrac{5}{6}$ .
$\Rightarrow \dfrac{1}{3}x=\dfrac{5}{6}-2$
$\Rightarrow \dfrac{1}{3}x=-\dfrac{7}{6}$
In order to make the coefficient of x equal to 1, we will now multiply the entire equation by 3.
$\Rightarrow 3\dfrac{1}{3}x=3\left( -\dfrac{7}{6} \right)$
$\Rightarrow x=-\dfrac{7}{2}$
The solution of the equation is the value of the variable-x which will be obtained on solving the equation and we have calculated variable x equal to $-\dfrac{7}{2}$ .

Therefore, the solution of the given equation $\dfrac{1}{3}x+2=\dfrac{5}{6}$ is $x=-\dfrac{7}{2}$.

Note: Another method of solving this equation was by transposing the constant term $\dfrac{5}{6}$ to the left hand side of the equation and subtracting it from 2 after which we would have got $\dfrac{1}{3}x+\dfrac{7}{6}=0$. Then we shall group the common factors, that is, $\dfrac{1}{3}$ in this case and separate the other terms as $\dfrac{1}{3}\left( x+\dfrac{7}{2} \right)=0$. Further we shall equate $\left( x+\dfrac{7}{2} \right)$ equal to 0 to compute the value of variable-x equal to $-\dfrac{7}{2}$.