Question

Solve for x in the below mentioned linear equation.
\[2x-3=\dfrac{3}{10}(5x-12)\]

Answer 1

Hint: Given an equation to solve is a linear equation in one variable (in x). Above equation can easily be solved just by using simple mathematical operations. First step is to separate variables and constant terms and then proceed step by step. Don’t forget to check whether found x value is correct or not.

Complete step-by-step answer:
The given expression is:
\[2x-3=\dfrac{3}{10}(5x-12)\]
First, we’ll separate variable coefficients and constants. Variable coefficients can be kept on L.H.S. or R.H.S. But here, variable coefficients are being placed on L.H.S.
We can see that the R.H.S. expression after multiplying by \[\dfrac{3}{10}\] will become:
\[\dfrac{3}{10}\times 5x-\dfrac{3}{10}\times 12\]
So, now we have the overall expression as shown below:
\[2x-3=\dfrac{3}{10}\times 5x-\dfrac{3}{10}\times 12\]
\[\Rightarrow 2x-3=\dfrac{3}{2}x-\dfrac{3}{5}\times 6\]
\[\Rightarrow 2x-3=\dfrac{3}{2}x-\dfrac{18}{5}\]
\[\Rightarrow 2x-\dfrac{3}{2}x=3-\dfrac{18}{5}\] …… (1)
Our strategy of separating variable coefficients & different constants has completed in equation (1).
Now, let’s perform basic mathematical operations to find out the value of x.
From equation (1), we have got already:
\[\Rightarrow 2x-\dfrac{3}{2}x=3-\dfrac{18}{5}\]
Now, taking x common from L.H.S. & taking L.C.M. on R.H.S. side, we get:
\[\left( 2-\dfrac{3}{2} \right)x=\left( \dfrac{3\times 5-18}{5} \right)\]
\[\Rightarrow \left( \dfrac{4-3}{2} \right)x=\left( \dfrac{3\times 5-18}{5} \right)\]
\[\Rightarrow \left( \dfrac{x}{2} \right)=\left( \dfrac{3\times 5-18}{5} \right)\]
\[\Rightarrow \left( \dfrac{x}{2} \right)=\left( \dfrac{15-18}{5} \right)\]
\[\Rightarrow \left( \dfrac{x}{2} \right)=\left( \dfrac{-3}{5} \right)\]
On cross multiplying the above equation, we get:
\[\Rightarrow x=2\times \left( \dfrac{-3}{5} \right)\]
\[\Rightarrow x=\left( \dfrac{-6}{5} \right)\]
Hence, the value of x is \[\left( \dfrac{-6}{5} \right)\].

Note: We can check whether we have got the correct value of x or not. This can be done just by simply assigning value of x in question and can check L.H.S. and R.H.S. If L.H.S. = R.H.S. Then, x value is correct, if L.H.S. is not equal to R.H.S. Then, x value is incorrect. To check whether the found value of x is correct or not. We’ll check L.H.S. & R.H.S. as explained below:
L.H.S. =\[2x-3\]
\[=2\left( \dfrac{-6}{5} \right)-3\]
\[=\left( \dfrac{-12-3\times 5}{5} \right)\]
\[=-\dfrac{27}{5}\]
So, L.H.S. \[=-\dfrac{27}{5}\]
Now, let us find R.H.S. value.
Since, R.H.S. =\[\dfrac{3}{10}(5x-12)\]
\[=\dfrac{3}{10}(5\times \left( \dfrac{-6}{5} \right)-12)\]
\[=\dfrac{3}{10}(-6-12)\]
\[=\dfrac{-27}{5}\]
So, L.H.S. = R.H.S.
Hence, x value is correct.