Question

How do you solve $\dfrac{1}{8}x + \dfrac{3}{2} = \dfrac{3}{4}x - 1$?

Answer 1

Hint: In this question we have to solve the equation for$x$, the given equation is a linear equation as the degree of the highest exponent of $x$ is equal to 1. To solve the equation first open the brackets then take all $x$ terms to one side and all constants to the others side and solve for required $x$.

Complete step by step answer:
Given equation is $\dfrac{1}{8}x + \dfrac{3}{2} = \dfrac{3}{4}x - 1$, and we have to solve for$x$,
Now subtract $\dfrac{3}{2}$ from both sides of the equation, we get,
$ \Rightarrow \dfrac{1}{8}x + \dfrac{3}{2} - \dfrac{3}{2} = \dfrac{3}{4}x - 1 - \dfrac{3}{2}$,
Now simplifying by eliminating the like terms, we get,
$ \Rightarrow \dfrac{1}{8}x = \dfrac{3}{4}x - \left( {\dfrac{{2 + 3}}{2}} \right)$,
Now simplifying we get,
$ \Rightarrow \dfrac{1}{8}x = \dfrac{3}{4}x - \dfrac{5}{2}$
Now subtract $\dfrac{3}{4}x$ to both sides of the equation we get,
$ \Rightarrow \dfrac{1}{8}x - \dfrac{3}{4}x = \dfrac{3}{4}x - \dfrac{3}{4}x - \dfrac{5}{2}$,
Now simplifying by L.C.M on the left hand sidewe get,
$ \Rightarrow \dfrac{1}{8}x - \dfrac{{3 \times 2}}{{4 \times 2}}x = - \dfrac{5}{2}$,
Now simplifying we get,
$ \Rightarrow \dfrac{1}{8}x - \dfrac{6}{8}x = - \dfrac{5}{2}$,
Now as denominators are equal simplify the expression,
$ \Rightarrow \left( {\dfrac{{1 - 6}}{8}} \right)x = - \dfrac{5}{2}$,
Now simplifying we get,
$ \Rightarrow - \dfrac{5}{8}x = - \dfrac{5}{2}$,
Now divide both sides with -5 we get,
$ \Rightarrow \dfrac{{ - \dfrac{5}{8}x}}{{ - 5}} = \dfrac{{ - \dfrac{5}{2}}}{{ - 5}}$,
Now simplifying we get,
$ \Rightarrow \dfrac{1}{8}x = \dfrac{1}{2}$,
Now multiply both sides with 8 we get,
$ \Rightarrow \dfrac{1}{8}x \times 8 = \dfrac{1}{2} \times 8$,
Now simplifying we get,
$ \Rightarrow x = 4$
So the value of $x$ will be 4, i.e., when we substitute the value of $x$ in the equation $\dfrac{1}{8}x + \dfrac{3}{2} = \dfrac{3}{4}x - 1$, then right hand side of the equation will be equal to left hand side of the equation, we get,
$ \Rightarrow \dfrac{1}{8}x + \dfrac{3}{2} = \dfrac{3}{4}x - 1$,
Now substitute $x = 4$, we get,
$ \Rightarrow \dfrac{1}{8}\left( 4 \right) + \dfrac{3}{2} = \dfrac{3}{4}\left( 4 \right) - 1$,
Now simplifying we get,
$ \Rightarrow \dfrac{1}{2} + \dfrac{3}{2} = 3 - 1$ ,
Further simplifying we get,
$ \Rightarrow \dfrac{1}{2} + \dfrac{3}{2} = 3 - 1$,
Now multiplying we get,
$ \Rightarrow \dfrac{{3 + 1}}{2} = 2$,
Now simplifying we get,
$ \Rightarrow \dfrac{4}{2} = 2$,
Now simplifying we get,
$ \Rightarrow 2 = 2$,
Therefore here R.H.S=L.H.S.

So, the value of $x$ is $4$.

Note: A linear equation is an equation of a straight line having a maximum of one variable. The degree of the variable will be equal to 1. To solve any equation in one variable, pit all the variable terms on the left hand side and all the numerical values on the right hand side to make the calculation solved easily.