Question

What is the value of $x$ in the given equation? $\dfrac{{3x + 1}}{{16}} + \dfrac{{2x - 3}}{7} = \dfrac{{x + 3}}{8} + \dfrac{{3x - 1}}{{14}}$

Answer 1

Hint: Here, we have to find the value of $x$ in the given equation. The given equation is a linear equation in one variable. Linear equations are the equations when we have a variable of maximum one order or degree in an equation. We will find the value of $x$ by solving the above equation by rearranging and taking L.C.M and further solving the equation we get our required result.

Complete step by step answer:
Here, we have to find the value of $x$ in the given equation. The given equation is a linear equation in one variable. Linear equations are the equations when we have a variable of maximum one order or degree in an equation. Here, the degree of $x$ is $1$ hence, the above equation is a linear equation in one variable.Now, we will find the value of $x$.We have,
$\dfrac{{3x + 1}}{{16}} + \dfrac{{2x - 3}}{7} = \dfrac{{x + 3}}{8} + \dfrac{{3x - 1}}{{14}}$

Rearranging the above equation. We get,
$ \Rightarrow \dfrac{{2x - 3}}{7} - \dfrac{{3x - 1}}{{14}} = \dfrac{{x + 3}}{8} - \dfrac{{3x + 1}}{{16}}$
Taking L.C.M of the denominator. The L.C.M of $7,14 = 14$ and the L.C.M of $8,16 = 16$. So,
$ \Rightarrow \dfrac{{2(2x - 3) - (3x - 1)}}{{14}} = \dfrac{{2(x + 3) - (3x + 1)}}{{16}}$
Solving the above equation. We get,
$ \Rightarrow \dfrac{{4x - 6 - 3x + 1}}{{14}} = \dfrac{{2x + 6 - 3x - 1}}{{16}}$
Solving $x$ terms and constant terms on both sides of the equation. We get,
$ \Rightarrow \dfrac{{x - 5}}{{14}} = \dfrac{{5 - x}}{{16}}$
$ \Rightarrow 16(x - 5) = 14(5 - x)$

On multiplying. We get,
$ \Rightarrow 16x - 80 = 70 - 14x$
Shifting $14x$ to the left side of the equation and $80$ to the right side of the equation. We get,
$ \Rightarrow 16x + 14x = 70 + 80$
Solving $x$ terms and constant terms of the equation. We get,
$ \Rightarrow 30x = 150$
Dividing $150$ by $30$. We get,
$ \Rightarrow x = \dfrac{{150}}{{30}}$
On solving we get,
$ \therefore x = 5$

Hence, the value of $x = 5$ in the given linear equation.

Note: Linear equations in one variable can be defined as when we have a variable of a maximum of one order in an equation and generally they can be expressed in the form of $jx + k = 0$ where $j$ and $k$ are constant terms and $x$ is the variable whose solution is only one as it has one degree in the equation. Note that the number of solutions of a variable depends on the degree of the variable.