Question

How do you solve the system of linear equation:
$\left( a+2b \right)x+\left( 2a-b \right)y=2$and $\left( a-2b \right)x+\left( 2a+b \right)y=3?$

Answer 1

Hint: Solve the given equation for any one of the variables. Then substitute that equation into the other equation and solve it for $x$ then substitute the value of $x$ in original equation and then solve it for $y.$ Sometimes there are $2$ equations which are easy to solve, but sometimes there are three equations which are a little bit trickier which is difficult to solve.

Complete step by step solution:
We have,
The given equation,
$\left( a+2b \right)x+\left( 2a-b \right)y=0...(i)$
$\left( a-2b \right)x+\left( 2a+b \right)y=3...(ii)$
By adding the equation $(i)$ and $(ii)$ we get
$2ax+4ay=5...(iii)$
And subtract the equation $(ii)$ from $(i)$
We get,
$4bx-2by=-1..(iv)$
Multiply $(iii)$ by $2b$ and $(iv)$ by $'a'$ we get,
$4abx+8aby=10b...(v)$
And
$4abx-2aby=-a...(vi)$
After that,
Subtracting $(vi)$ from $(v)$ we have,
$10aby=10b+a$ or $y=\dfrac{10b+a}{10ab}=\dfrac{1}{a}+\dfrac{1}{10b}$
Multiplying $(vi)$ by $(iv)$ and adding to $(v)$ we get,
$200abx=10b-4a$ or
$x=\dfrac{10b-4a}{20ab}=\dfrac{1}{2a}-\dfrac{1}{5b}$

Additional Information:
The method of solving by substitution works by solving one of the equations for one of the variables and then adding this back into another equation substituting for the chosen variable and solving for another. Then again solving the given question has given you the idea to solve one of the equations for the one of the variables, and plug this into another question. It does not matter which equation or variable you choose. There is no right or wrong choice. The answer you get will be the same.

Note: When you are solving the problem check the given value. Then you get the idea of what we have to find. Firstly you have to simplify the given equation by expanding. After that check the step whether you have solved the correct variable or not and substitute the proper values analyse. The problem totally checks all the possibilities where you get wrong in the problem. Re-calculate the values sometime it is misplaced due to the answer getting incorrect.