Question

In the given equations solve for x and y:
\[\left[ \begin{matrix}
   \left( a+2b \right)x+\left( 2a-b \right)y=2 \\
   \left( a-2b \right)x+\left( 2a+b \right)y=3 \\
\end{matrix} \right]\]

Answer 1

Hint: Apply the substitution method. Find x in terms of y from the first equation\[\left( a+2b \right)x+\left( 2a-b \right)y=2\], then replace this x in the second equation\[\left( a-2b \right)x+\left( 2a+b \right)y=3\], to get the equation in just y. Then solve for y then get the corresponding value of x.

Complete step by step answer:
In the question, we have to solve the set of linear equations \[\left[ \begin{matrix}
   \left( a+2b \right)x+\left( 2a-b \right)y=2 \\
   \left( a-2b \right)x+\left( 2a+b \right)y=3 \\
\end{matrix} \right]\]to get the value of x and y. So we will use the substitution method here.
So let the first equation is \[\left( a+2b \right)x+\left( 2a-b \right)y=2\], which we will use to find the value of x in terms of y, as shown below:
\[\begin{align}
  & \Rightarrow \left( a+2b \right)x+\left( 2a-b \right)y=2 \\
 & \Rightarrow \left( a+2b \right)x=2-\left( 2a-b \right)y \\
 & \Rightarrow x=\dfrac{2-y\left( 2a-b \right)}{a+2b}\,,\,\,\,where\,\,a\ne \;-2b \\
\end{align}\]
Now, the above obtained value of x has to be put in the second equation that is \[\left( a-2b \right)x+\left( 2a+b \right)y=3\], and we will get:
\[\begin{align}
  & \Rightarrow \left( a-2b \right)x+\left( 2a+b \right)y=3 \\
 & \Rightarrow \left( a-2b \right)\dfrac{2-y\left( 2a-b \right)}{a+2b}+\left( 2a+b \right)y=3 \\
\end{align}\]
Now, we will further simplify by taking the common denominator.
So, we get:
\[\begin{align}
  & \Rightarrow \dfrac{\left( 2-y\left( 2a-b \right) \right)\left( a-2b \right)}{a+2b}+y\left( 2a+b \right)=3 \\
 & \Rightarrow \dfrac{\left( 2-y\left( 2a-b \right) \right)\left( a-2b \right)}{a+2b}+\dfrac{2ay\left( a+2b \right)}{a+2b}+\dfrac{by\left( a+2b \right)}{a+2b}=3 \\
 & \Rightarrow \dfrac{\left( 2-y\left( 2a-b \right) \right)\left( a-2b \right)+2ay\left( a+2b \right)+by\left( a+2b \right)}{a+2b}=3 \\
\end{align}\]
Next, we will simplify the numerator and then cross multiply to get the y, as shown below:
\[\begin{align}

 & \Rightarrow \dfrac{10ab\,y+2a-4b}{a+2b}=3 \\
 & \Rightarrow 10aby+2a-4b=3\left( a+2b \right) \\
 & \Rightarrow \text{10aby=a+10b} \\
\end{align}\]
After dividing both sides by 10ab. Now, we will further simplify after cancelling out the common terms, as follows:
\[\begin{align}
  & \Rightarrow \dfrac{10aby}{10ab}=\dfrac{a}{10ab}+\dfrac{10b}{10ab} \\
 & \Rightarrow y=\left( \dfrac{1}{10b}+\dfrac{1}{a} \right) \\
\end{align}\]

Next, from this y value, we will find the value of x using the result \[x=\dfrac{2-y\left( 2a-b \right)}{a+2b}\]
So the value of x will be found as follows:
\[\begin{align}
  & \Rightarrow x=\dfrac{2-y\left( 2a-b \right)}{a+2b} \\
 & \Rightarrow x=\dfrac{2-\left( \dfrac{1}{10b}+\dfrac{1}{a} \right)\left( 2a-b \right)}{a+2b} \\
\end{align}\]

Now, here we will take the common denominator and further simplify it, as shown below:
\[\begin{align}
 & \Rightarrow x=\dfrac{2-\dfrac{\left( a+10b \right)\left( 2a-b \right)}{10ab}}{a+2b} \\
 & \Rightarrow x=\dfrac{\left( a+2b \right)\left( -2a+5b \right)}{10ab\left( a+2b \right)} \\
 & \Rightarrow x=\dfrac{-2a+5b}{10ab} \\
 & \Rightarrow x=\dfrac{-2a}{10ab}+\dfrac{5b}{10ab} \\

\end{align}\]
Now, separating the right-hand side and cancelling out the common terms, we get:
\[\begin{align}
  & \Rightarrow x=\dfrac{-2a}{10ab}+\dfrac{5b}{10ab} \\
 & \Rightarrow x=-\dfrac{1}{5b}+\dfrac{1}{2a} \\
\end{align}\]
This gives the value of x.

So the required value of x and y are as follows:
\[x=\dfrac{1}{2a}-\dfrac{1}{5b}\] and \[y=\left( \dfrac{1}{10b}+\dfrac{1}{a} \right)\]

Note: The alternate way to solve this equation would be the elimination method. Where we will make the coefficients of x same in the two given equations, then eliminate the variable x. Now, there will be just one variable y is left and we can easily solve for y. Next, put the value of y in any equation to get the corresponding value of x.