Question

Solve the equation \[\dfrac{{3x - 1}}{2} = \dfrac{{2x + 6}}{3}\]?

Answer 1

Hint: We have an equation with one variable. Here we need to solve for ‘x’. First, we cross multiply it. After that, we use the transposition method to simplify further. That is grouping the variable on one side of the equation and constant on the other side of the equation.

Complete step-by-step solution:
Given, \[\dfrac{{3x - 1}}{2} = \dfrac{{2x + 6}}{3}\].
Cross multiplying, we have,
\[3\left( {3x - 1} \right) = 2\left( {2x + 6} \right)\]
Expanding the brackets we have,
\[9x - 3 = 4x + 12\]
We transpose \[ - 3\] which is present in the left-hand side of the equation to the right-hand side of the equation by adding \[3\] on the right-hand side of the equation.
\[9x = 4x + 12 + 3\]
Similarly, we transpose 4x to the right-hand side or the equation by subtracting 4x on the LHS.
\[9x - 4x = 12 + 3\]
We can see that the variable ‘x’ and the constants are separated, then
\[5x = 15\]
We divide the whole equation by 5,
\[x = \dfrac{{15}}{5}\]
\[ \Rightarrow x = 3\].
This is the required result.

Note: We can check whether the obtained solution is correct or wrong. All we need to do is substituting the value of ‘x’ in the given problem.
\[\dfrac{{3x - 1}}{2} = \dfrac{{2x + 6}}{3}\]
\[\dfrac{{3(3) - 1}}{2} = \dfrac{{2(3) + 6}}{3}\]
\[\dfrac{{9 - 1}}{2} = \dfrac{{6 + 6}}{3}\]
\[\dfrac{8}{2} = \dfrac{{12}}{3}\]
\[ \Rightarrow 4 = 4\]
Hence the obtained answer is correct.
In the above we did the transpose of addition and subtraction. Similarly, if we have multiplication, we use division to transpose. If we have division, we use multiplication to transpose. Follow the same procedure for these kinds of problems.