Question

Solve the following pairs of equations by reducing them to a pair of linear equations.
\[\dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2\] and \[\dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6}\]

Answer 1

Hint: We use substitution of reciprocal of the given variables as two new different variables and solve the two equations formed using new variables. Solve the linear equations formed using substitution method.
* Substitution method: Use values of variables obtained using one of the equations and substitute that value in another equation to calculate the value of the variable.

Complete step-by-step solution:
We are given two equations \[\dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2\] and \[\dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6}\]..................… (1)
Since the two variables in the equations are ‘x’ and ‘y’; we use the substitution of reciprocals of ‘x’ and ‘y’.
Substitute \[\dfrac{1}{x} = a\] and \[\dfrac{1}{y} = b\] in equation (1)
Equations become:
\[ \Rightarrow \]\[\dfrac{a}{2} + \dfrac{b}{3} = 2\] and \[\dfrac{a}{3} + \dfrac{b}{2} = \dfrac{{13}}{6}\]
Now we take LCM in each of the given equations in LHS of the equations
\[ \Rightarrow \]\[\dfrac{{3a + 2b}}{{2 \times 3}} = 2\] and \[\dfrac{{2a + 3b}}{{3 \times 2}} = \dfrac{{13}}{6}\]
Cross multiply values from denominator of LHS to RHS of the equations
\[ \Rightarrow \]\[3a + 2b = 2 \times 6\] and \[2a + 3b = \dfrac{{13}}{6} \times 6\]
Cancel same factors from numerator and denominator in RHS of the second equation
\[ \Rightarrow \]\[3a + 2b = 12\] and \[2a + 3b = 13\]..................… (2)
Now we system of linear equations in equation (2)
We have \[3a + 2b = 12\]
Shift 2b to RHS of the equation
\[ \Rightarrow 3a = 12 - 2b\]
Divide both sides of equation by 3
\[ \Rightarrow a = \dfrac{{12 - 2b}}{3}\].................… (3)
Substitute the value of ‘a’ in equation \[2a + 3b = 13\]
\[ \Rightarrow 2\left( {\dfrac{{12 - 2b}}{3}} \right) + 3b = 13\]
Multiply terms outside the bracket with the fraction in LHS
\[ \Rightarrow \dfrac{{2 \times 12 - 2 \times 2b}}{3} + 3b = 13\]
\[ \Rightarrow \dfrac{{24 - 4b}}{3} + 3b = 13\]
Take LCM in LHS of the equation
\[ \Rightarrow \dfrac{{24 - 4b + 9b}}{3} = 13\]
Cross multiply the values from LHS to RHS
\[ \Rightarrow 24 + 5b = 13 \times 3\]
\[ \Rightarrow 24 + 5b = 39\]
Shift all constant values to RHS of the equation
\[ \Rightarrow 5b = 39 - 24\]
\[ \Rightarrow 5b = 15\]
Divide both sides by 5
\[ \Rightarrow \dfrac{{5b}}{5} = \dfrac{{15}}{5}\]
Cancel same terms from numerator and denominator on both sides of the equation
\[ \Rightarrow b = 3\]
Since we had substitution \[\dfrac{1}{y} = b\]
\[ \Rightarrow \dfrac{1}{y} = 3\]
Taking reciprocal on both sides
\[ \Rightarrow y = \dfrac{1}{3}\]................… (4)
Now substitute the value of b in equation (3) to calculate ‘a’
\[ \Rightarrow a = \dfrac{{12 - 2 \times 3}}{3}\]
\[ \Rightarrow a = \dfrac{{12 - 6}}{3}\]
\[ \Rightarrow a = \dfrac{6}{3}\]
Cancel same factors from numerator and denominator on right hand side of the equation
\[ \Rightarrow a = 2\]
Since we had substitution \[\dfrac{1}{x} = a\]
\[ \Rightarrow \dfrac{1}{x} = 2\]
Taking reciprocal on both sides
\[ \Rightarrow x = \dfrac{1}{2}\]................… (5)

\[\therefore \]Solution of system of equations \[\dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2\] and \[\dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6}\] is \[x = \dfrac{1}{2};y = \dfrac{1}{3}\]

Note: Many students make mistakes solving the given equations directly by taking LCM and then try to remove the term with variable ‘xy’ which will give us only one equation with variables ‘x’ and ‘y’. Keep in mind that we need two linear equations to solve the system of equations.