Question

Solve the given equation \[\dfrac{{2x + 7}}{5} - \dfrac{{3x + 11}}{2} = \dfrac{{2x + 8}}{3} - 5\]?

Answer 1

Hint: In the given problem we need to solve this for ‘x’. We can solve this using the transposition method. The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolating the variable (or the unknown quantity). That is we group the ‘x’ terms one side and constants on the other side of the equation.

Complete step-by-step answer:
Given,
\[ \Rightarrow \,\,\dfrac{{2x + 7}}{5} - \dfrac{{3x + 11}}{2} = \dfrac{{2x + 8}}{3} - 5\]
To simplify, take L.C.M for 2 and 5 is 10 in LHS and 3 in RHS, then
\[ \Rightarrow \,\,\dfrac{{2\left( {2x + 7} \right) - 5\left( {3x + 11} \right)}}{{10}} = \dfrac{{2x + 8 - 15}}{3}\]
On simplification, we have
\[ \Rightarrow \,\,\dfrac{{4x + 14 - 15x - 55}}{{10}} = \dfrac{{2x - 7}}{3}\]
\[ \Rightarrow \,\,\dfrac{{ - 11x - 41}}{{10}} = \dfrac{{2x - 7}}{3}\]
On cross multiplication, we have
\[ \Rightarrow \,\,3\left( { - 11x - 41} \right) = 10\left( {2x - 7} \right)\]
Remove the parenthesis by multiplying the outer number, then
\[ \Rightarrow \,\, - 33x - 123 = 20x - 70\]
We transpose ‘-20x’ which is present in the right hand side of the equation to the left hand side of the equation by subtracting ‘20x’ on the left hand side of the equation.
\[ \Rightarrow \,\, - 33x - 123 - 20x = - 70\]
\[ \Rightarrow \,\, - 53x - 123 = - 70\]
Transpose -123 to the right hand side of the equation by adding 123 on the right hand side of the equation,
\[ \Rightarrow \,\, - 53x = - 70 + 123\]
\[ \Rightarrow \,\, - 53x = 53\]
Now dividing ‘-53’ on both side of the equation,
\[ \Rightarrow \,\,x = \dfrac{{53}}{{ - 53}}\]
\[ \Rightarrow x = - 1\]
This is the required answer.

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.
\[ \Rightarrow \,\,\dfrac{{2\left( { - 1} \right) + 7}}{5} - \dfrac{{3\left( { - 1} \right) + 11}}{2} = \dfrac{{2\left( { - 1} \right) + 8}}{3} - 5\]
\[ \Rightarrow \,\,\dfrac{{ - 2 + 7}}{5} - \dfrac{{ - 3 + 11}}{2} = \dfrac{{ - 2 + 8}}{3} - 5\]
\[ \Rightarrow \,\,\dfrac{5}{5} - \dfrac{8}{2} = \dfrac{6}{3} - 5\]
\[ \Rightarrow \,\,1 - \dfrac{8}{2} = \dfrac{6}{3} - 5\]
Take L.C.M 2 in LHS and 3 in RHS, then
\[ \Rightarrow \,\,\dfrac{{2 - 8}}{2} = \dfrac{{6 - 15}}{3}\]
\[ \Rightarrow \,\,\dfrac{{ - 6}}{2} = \dfrac{{ - 9}}{3}\]
\[ \Rightarrow \,\, - 3 = - 3\]
Hence the obtained answer is correct.
We know that the product of two negative numbers is a positive number. Product of a negative number and a positive number gives negative number (vice versa). If we want to transpose a positive number to the other side of the equation we subtract the same number on that side (vice versa). 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.