Question

Solve the following equations:
\[\dfrac{1}{2\left( 2x+3y \right)}+\dfrac{12}{7\left( 3x-2y \right)}=\dfrac{1}{2}\] and
\[\dfrac{7}{2x+3y}+\dfrac{4}{3x-2y}=2\], where \[2x+3y\ne 0\] and \[3x-2y\ne 0\].

Answer 1

Hint: In this question, let us first assume the denominators as some other variables. Then solve both the equations by using the method of elimination by equating the coefficients to get the variable values. Then substitute back the variable values and simplify further again to get the values of x and y.

Complete step-by-step answer:
Linear equation: An equation involving the variables in maximum of order 1 is called a linear equation.
Linear equation in two variables is of the form \[ax+by+c=0\]
Now, from the given equations in the question we have
\[\begin{align}
  & \dfrac{1}{2\left( 2x+3y \right)}+\dfrac{12}{7\left( 3x-2y \right)}=\dfrac{1}{2} \\
 & \dfrac{7}{2x+3y}+\dfrac{4}{3x-2y}=2 \\
\end{align}\]
Now, let us assume as follows
\[\begin{align}
  & \dfrac{1}{2x+3y}=X \\
 & \dfrac{1}{3x-2y}=Y \\
\end{align}\]
Now, let us substitute back these assumed values in the given equations then we get,
\[\Rightarrow \dfrac{X}{2}+\dfrac{12Y}{7}=\dfrac{1}{2}\]
Now, let us take the L.C.M on left hand side to simplify it further
\[\Rightarrow \dfrac{7X+24Y}{7\times 2}=\dfrac{1}{2}\]
Let us now cancel the common terms on both the sides and cross multiply
\[\Rightarrow 7X+24Y=7......\left( 1 \right)\]
Now, let us consider the other equation and substitute the assumed values
\[\Rightarrow 7X+4Y=2......\left( 2 \right)\]
Now, let us further solve the equations (1) and (2)
\[\begin{align}
  & \Rightarrow 7X+24Y=7 \\
 & \Rightarrow 7X+4Y=2 \\
\end{align}\]
Now, on subtracting these two equations we get,
\[\Rightarrow 24Y-4Y=7-2\]
Now, on simplifying it further we get,
\[\Rightarrow 20Y=5\]
Let us now divide with 20 on both the sides
\[\therefore Y=\dfrac{1}{4}\]
Let us now substitute back this value in equation (2)
\[\Rightarrow 7X+4\times \dfrac{1}{4}=2\]
Now, on rearranging the terms we get,
\[\Rightarrow 7X=2-1\]
Let us now divide with 7 on both the sides
\[\therefore X=\dfrac{1}{7}\]
Now, from the assumed values we have
\[\Rightarrow \dfrac{1}{2x+3y}=X\]
Now, on substituting the value we get,
\[\Rightarrow \dfrac{1}{2x+3y}=\dfrac{1}{7}\]
Now, on further simplification we get,
\[\Rightarrow 2x+3y=7....\left( 3 \right)\]
Now, on considering the other equation and from the assumed values we get,
\[\Rightarrow \dfrac{1}{3x-2y}=Y\]
Now, on substituting the respective value we get,
\[\Rightarrow \dfrac{1}{3x-2y}=\dfrac{1}{4}\]
 Now, on further simplification we get,
\[\Rightarrow 3x-2y=4....\left( 4 \right)\]
Now, from the above equations (3) and (4) we have
\[\begin{align}
  & \Rightarrow 2x+3y=7 \\
 & \Rightarrow 3x-2y=4 \\
\end{align}\]
Now, let us multiply equation (3) with 2 and equation (4) with 3 then we get,
\[\begin{align}
  & \Rightarrow 4x+6y=14 \\
 & \Rightarrow 9x-6y=12 \\
\end{align}\]
Now, let us add these to equations to simplify further
\[\Rightarrow 4x+9x=14+12\]
Now, let us simplify it further
\[\Rightarrow 13x=26\]
Let us now divide with 13 on both sides
\[\therefore x=2\]
Let us now substitute this value of x in equation (3)
\[\Rightarrow 2\times 2+3y=7\]
Now, on rearranging the terms we get,
\[\Rightarrow 3y=7-4\]
\[\therefore y=1\]
Hence, we get the values of x as 2 and y as 1

Note: Instead of assuming some variables to the denominators and then finding their values we can also directly solve by first cross multiplying the equations given and then rearrange the terms. Then on further simplification we get the result.
It is important to note that finding the assumed values is not the final result. We then need to solve the equations accordingly without neglecting any of the terms and find the values of x and y.