Question

Solve for the value of x: \[(x - 5)/2 - (x - 3)/5 = 1/2\]

Answer 1

Hint: We use LCM to solve the LHS of the equation first and then shifting values to suitable sides make an equation where variable is on one side and constant value is on other. We find the value of the variable in the equation and as there is only one unknown variable, i.e. ‘x’ we evaluate the equation for the value of ‘x’.

Complete step-by-step solution:
We have to solve the equation \[\dfrac{{(x - 5)}}{2} - \dfrac{{(x - 3)}}{5} = \dfrac{1}{2}\] ………..… (1)
Since here the variable is ‘x’ in the equation, we will find the value of ‘x’ using operations like multiplication, addition, subtraction and division.
Solve LHS of the equation (1)
Take LCM of 2 and 5 in LHS of the equation
\[ \Rightarrow \dfrac{{5(x - 5) - 2(x - 3)}}{{2 \times 5}} = \dfrac{1}{2}\]
Calculate products in numerator of LHS
\[ \Rightarrow \dfrac{{5x - 25 - 2x + 6}}{{2 \times 5}} = \dfrac{1}{2}\]
Cancel same factors from both sides of the denominator
\[ \Rightarrow \dfrac{{5x - 25 - 2x + 6}}{5} = 1\]
Pair the terms with variable and constants separately in numerator in LHS
\[ \Rightarrow \dfrac{{\left( {5x - 2x} \right) + \left( {6 - 25} \right)}}{5} = 1\]
\[ \Rightarrow \dfrac{{3x - 19}}{5} = 1\]
Cross multiply the value from denominator in LHS to RHS
\[ \Rightarrow 3x - 19 = 5\]
Shift all constants to right hand side of the equation
\[ \Rightarrow 3x = 5 + 19\]
Calculate the sum in right hand side of the equation
\[ \Rightarrow 3x = 24\]
Divide both sides of the equation by 3
\[ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{24}}{3}\]
Cancel same factors from numerator and denominator on both sides of the equation
\[ \Rightarrow x = 8\]
So, the value of \[x = 8\] for the equation \[(x - 5)/2 - (x - 3)/5 = 1/2\]

\[\therefore \]Solution of the equation \[(x - 5)/2 - (x - 3)/5 = 1/2\] is \[x = 8\]

Note: Students are likely to make mistakes in the calculation part where we shift values from one side of the equation to another, so always keep in mind that the sign changes from positive to negative and vice versa when shifting any constant or variable to the other side.