Question

If \[2y + \dfrac{5}{3} = \dfrac{{26}}{3} - y\] , then find value of \[y = \]
A.\[1\]
B.\[\dfrac{2}{3}\]
C.\[\dfrac{6}{5}\]
D.\[\dfrac{7}{3}\]

Answer 1

Hint: Here, we have to solve the given equation for \[y\]. The given equation is a linear equation in one variable. Linear equation is an equation which has a highest degree of the variable of 1. We will subtract the like terms and divide.

Complete step-by-step answer:
We are given a linear equation in one variable \[2y + \dfrac{5}{3} = \dfrac{{26}}{3} - y\].
Adding \[y\]on both the sides, we get
\[ \Rightarrow 2y + y + \dfrac{5}{3} = \dfrac{{26}}{3}\]
Now subtracting \[\dfrac{5}{3}\] from both the sides, we get
\[ \Rightarrow 2y + y = \dfrac{{26}}{3} - \dfrac{5}{3}\]
Like fractions are fractions whose numerators are different whereas the denominators are the same. Since the fraction is a like fraction, we can subtract the numerators.
Adding and subtracting the like terms, we get
\[ \Rightarrow 3y = \dfrac{{21}}{3}\]
Simplifying the equation, we get
\[ \Rightarrow 3y = 7\]
Dividing the equation by \[3\] , we have
\[ \Rightarrow y = \dfrac{7}{3}\]
Therefore, \[y = \dfrac{7}{3}\]

Note: We can solve any linear equation in one variable. Both sides of the equation need to be balanced for solving a linear equation. Equality sign denotes that the expressions on either side of the ‘equal to’ sign are equal. We need to note that the highest power of the variable in these expressions is 1.
Every linear equation in one variable has only one unique solution. Similarly, there are linear equations in two variables, linear equations in three variables and so on. We can solve the linear equations in two variables using elimination and substitution methods and the linear equations in three variables by using the matrix method. Other than linear equations, there are other types of equations as well. They are quadratic equation, cubic equation, quartic equation and so on.