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How do you solve the following system of equations?
$2x + y = 8$ , $- 2x + 3y = 12$

Last updated date: 20th Jun 2024
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Hint: To solve two given linear equations simultaneously, try eliminating one variable from both the equations. Then find the value of the remaining variable and then by substituting its value in any one of the two given linear equations, find the value of the variable that was eliminated.

Complete step by step answer:
The given system of linear equations in two variables are
$2x + y = 8$ $...(i)$
$- 2x + 3y = 12$ $...(ii)$
We will try to solve these equations by the method of elimination, which is by eliminating one of the variables from both the equations.
In the given equations we will eliminate the variable $x$ and find the value of the other variable $y$ .
Add the above equations $(i)$ and $(ii)$ to get the L.H.S and R.H.S values as
$2x + y + \left( { - 2x + 3y} \right) = 8 + 12$
Collecting, combining, and cancelling the like terms on both the sides we get
$\Rightarrow 2x - 2x + 3y + y = 20$
$\Rightarrow 4y = 20$
Dividing both the sides of the above equation by $5$ to get
$\Rightarrow y = 5$
To find the value of the previously eliminated variable $x$ we need to substitute the value of $y$ in any of the one linear equations.
Therefore, substituting the value of $y$in $2x + y = 8$ that is the equation $(i)$ we get:
$2x + 5 = 8$
Taking the constants on the R.H.S of the linear equation
$\Rightarrow 2x = 8 - 5$
$\Rightarrow 2x = 3$
Dividing both the sides of the above expression by $2$ to get
$\Rightarrow x = \dfrac{3}{2}$

Hence, on solving the given pair of linear equations by the elimination method we get the values of the variables as $x = \dfrac{3}{2}$ and $y = 5$

Note: There are three main ways for solving simultaneous linear equations which are: Elimination by a variable, Substitution method, and Cross multiplication method. Therefore the above pair of linear equations can also be solved with the help of substitution and cross multiplication method. Also, you can check whether the calculated values of the variable are correct by substituting the value back in one of the equations.