Solve the pair of equations
\[ \dfrac{3}{{2x}} + \dfrac{2}{{3y}} = \dfrac{{ - 1}}{3} \\
\dfrac{3}{{4x}} + \dfrac{1}{{2y}} = \dfrac{{ - 1}}{8} \]
Answer
522.3k+ views
Hint: We will use the method of elimination by equating the coefficients of either \[x\] or \[y\]to find the desired solution that satisfies both the equations. Eliminate either of the variables and substitute it in one of the equations to get the other variable.
Complete step-by-step solution:
We are given with the equations,
\[\dfrac{3}{{2x}} + \dfrac{2}{{3y}} = \dfrac{{ - 1}}{3}\]--------- \[(1)\]
\[\dfrac{3}{{4x}} + \dfrac{1}{{2y}} = \dfrac{{ - 1}}{8}\]--------- \[(2)\]
Now multiplying equation \[(1)\] with \[\dfrac{1}{2}\] and equation \[(2)\] with \[1\] and subtracting we get,
\[\dfrac{1}{2} \times (\dfrac{3}{{2x}} + \dfrac{2}{{3y}} = \dfrac{{ - 1}}{3})\]
\[ \Rightarrow \dfrac{3}{{4x}} + \dfrac{1}{{3y}} = \dfrac{{ - 1}}{6}\]--------\[(3)\]
\[\dfrac{3}{{4x}} + \dfrac{1}{{2y}} = \dfrac{{ - 1}}{8}\]----------\[(4)\]
Now subtracting we get,
\[
\Rightarrow \dfrac{3}{{4x}} + \dfrac{1}{{3y}} + \dfrac{1}{6} - (\dfrac{3}{{4x}} + \dfrac{1}{{2y}} + \dfrac{1}{8}) = 0 \\
\Rightarrow \dfrac{1}{{3y}} - \dfrac{1}{{2y}} = \dfrac{1}{8} - \dfrac{1}{6} \\
\Rightarrow \dfrac{1}{y}(\dfrac{1}{3} - \dfrac{1}{2}) = \dfrac{6}{{8 \times 6}} - \dfrac{8}{{6 \times 8}} \\
\Rightarrow \dfrac{1}{y}(\dfrac{2}{{3 \times 2}} - \dfrac{3}{{2 \times 3}}) = \dfrac{{ - 2}}{{48}} \\
\Rightarrow \dfrac{{ - 1}}{{6y}} = \dfrac{{ - 2}}{{48}} \\
\Rightarrow \dfrac{1}{{6y}} = \dfrac{1}{{24}} \\
\Rightarrow 6y = 24 \\
\Rightarrow y = 4 \]
Now substituting \[y = 4\]in equation\[(1)\]we get,
\[
\dfrac{3}{{2x}} + \dfrac{2}{{3.4}} = \dfrac{{ - 1}}{3} \\
\Rightarrow \dfrac{3}{{2x}} + \dfrac{2}{{12}} = \dfrac{{ - 1}}{3} \\
\Rightarrow \dfrac{3}{{2x}} = - \dfrac{2}{{12}} - \dfrac{1}{3} \\
\Rightarrow \dfrac{3}{{2x}} = - (\dfrac{1}{6} + \dfrac{1}{3}) \\
\Rightarrow \dfrac{3}{{2x}} = - (\dfrac{1}{6} + \dfrac{2}{6}) \\
\Rightarrow \dfrac{3}{{2x}} = - \dfrac{3}{6} \\
\Rightarrow 2x = - 6 \\
\Rightarrow x = - 3 \]
Therefore the solution to the equations is \[( - 3,4)\]
Additional information: A linear equation is an equation that is written for two exceptional variables. This equation may be a linear aggregate of those variables, and a steady state can be present. Surprisingly, while any linear equation is plotted on a graph, it will necessarily produce an instantaneous line - as a result they are called: Linear equations. Linear Equations are an extensive sort of equations altogether. There may be linear equations in one variable, linear equations in two variables, and so on. In every equation, one issue stays regular: The highest and the best diploma of all variables in the equation should be \[1\]. Other than that, constants \[0\] diploma variables can be there.
Note: It is very important that we know how to calculate the LCM of the numbers if needed and then take one variable from either of the variables to calculate the LCM of the coefficients and then multiply each of them to eliminate either of the variables and then calculate the value of the other variable using substitution method.
Complete step-by-step solution:
We are given with the equations,
\[\dfrac{3}{{2x}} + \dfrac{2}{{3y}} = \dfrac{{ - 1}}{3}\]--------- \[(1)\]
\[\dfrac{3}{{4x}} + \dfrac{1}{{2y}} = \dfrac{{ - 1}}{8}\]--------- \[(2)\]
Now multiplying equation \[(1)\] with \[\dfrac{1}{2}\] and equation \[(2)\] with \[1\] and subtracting we get,
\[\dfrac{1}{2} \times (\dfrac{3}{{2x}} + \dfrac{2}{{3y}} = \dfrac{{ - 1}}{3})\]
\[ \Rightarrow \dfrac{3}{{4x}} + \dfrac{1}{{3y}} = \dfrac{{ - 1}}{6}\]--------\[(3)\]
\[\dfrac{3}{{4x}} + \dfrac{1}{{2y}} = \dfrac{{ - 1}}{8}\]----------\[(4)\]
Now subtracting we get,
\[
\Rightarrow \dfrac{3}{{4x}} + \dfrac{1}{{3y}} + \dfrac{1}{6} - (\dfrac{3}{{4x}} + \dfrac{1}{{2y}} + \dfrac{1}{8}) = 0 \\
\Rightarrow \dfrac{1}{{3y}} - \dfrac{1}{{2y}} = \dfrac{1}{8} - \dfrac{1}{6} \\
\Rightarrow \dfrac{1}{y}(\dfrac{1}{3} - \dfrac{1}{2}) = \dfrac{6}{{8 \times 6}} - \dfrac{8}{{6 \times 8}} \\
\Rightarrow \dfrac{1}{y}(\dfrac{2}{{3 \times 2}} - \dfrac{3}{{2 \times 3}}) = \dfrac{{ - 2}}{{48}} \\
\Rightarrow \dfrac{{ - 1}}{{6y}} = \dfrac{{ - 2}}{{48}} \\
\Rightarrow \dfrac{1}{{6y}} = \dfrac{1}{{24}} \\
\Rightarrow 6y = 24 \\
\Rightarrow y = 4 \]
Now substituting \[y = 4\]in equation\[(1)\]we get,
\[
\dfrac{3}{{2x}} + \dfrac{2}{{3.4}} = \dfrac{{ - 1}}{3} \\
\Rightarrow \dfrac{3}{{2x}} + \dfrac{2}{{12}} = \dfrac{{ - 1}}{3} \\
\Rightarrow \dfrac{3}{{2x}} = - \dfrac{2}{{12}} - \dfrac{1}{3} \\
\Rightarrow \dfrac{3}{{2x}} = - (\dfrac{1}{6} + \dfrac{1}{3}) \\
\Rightarrow \dfrac{3}{{2x}} = - (\dfrac{1}{6} + \dfrac{2}{6}) \\
\Rightarrow \dfrac{3}{{2x}} = - \dfrac{3}{6} \\
\Rightarrow 2x = - 6 \\
\Rightarrow x = - 3 \]
Therefore the solution to the equations is \[( - 3,4)\]
Additional information: A linear equation is an equation that is written for two exceptional variables. This equation may be a linear aggregate of those variables, and a steady state can be present. Surprisingly, while any linear equation is plotted on a graph, it will necessarily produce an instantaneous line - as a result they are called: Linear equations. Linear Equations are an extensive sort of equations altogether. There may be linear equations in one variable, linear equations in two variables, and so on. In every equation, one issue stays regular: The highest and the best diploma of all variables in the equation should be \[1\]. Other than that, constants \[0\] diploma variables can be there.
Note: It is very important that we know how to calculate the LCM of the numbers if needed and then take one variable from either of the variables to calculate the LCM of the coefficients and then multiply each of them to eliminate either of the variables and then calculate the value of the other variable using substitution method.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Which Indian city is known as the "City of Victory"?

Which instrument is used to measure the Blood Pressure?

Who Won 36 Oscar Awards? Record Holder Revealed

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

