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

Last updated date: 09th Aug 2024
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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
$x=\dfrac{10b-4a}{20ab}=\dfrac{1}{2a}-\dfrac{1}{5b}$