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# How do you solve the system using the elimination method for $3x+y=5$ and $x-2y=4?$

Last updated date: 19th Jun 2024
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Hint: Here two equations are given which we have to solve by using elimination method. ‘
In this method we have to make one of the coefficients or variables equal for both of the equations.
After making it equal, that coefficient or variable gets cancelled while performing adding or subtracting.

Complete step by step solution:
The given equation we have to solve by using elimination method.
The equation given are:
$3x+y=5...(i)$
$x-2y=4...(ii)$
Now, we have to make the $y$ coefficient equal for both the equations.
For that we have to multiply the equation $(i)$ by $2$ then we will get,
So,
$2\times \left( 3x+y=5 \right)$
$6x+2y=10...(iii)$
Now, the equation $(ii)$ and equation $(iii)$ have equal $y$ coefficient with opposite sign, So then it can be cancelled while adding.
So, add equation $(ii)$ and equation $(iii)$
$\left( x-2y=4 \right)+\left( 6x+2y=10 \right)$
$=7x=14$
$\Rightarrow x=\dfrac{14}{7}$
$\Rightarrow x=2$
Put, the $x=2$ in equation $(i)$
$3x+y=5$ (given)
$\Rightarrow 3\times 2+y=5$
$\Rightarrow6+y=5$
Transpose the $6$ to other side of equation
$y=5-6$
$\Rightarrow y=-1$

Therefore the solution of given equation by elimination method is $x=2$ and $y=-1$

There are so many methods of solving the system of equations.
But two of them are most preferable, one is elimination and second is substitution. The elimination method is that, in which we have to add or subtract the given equation for getting an answer of the equation in one variable. In the substitution method, from one equation, the other value of one variable is substituted in another equation.

Note: In equation $(iii)$ we multiply it by $2$ for making the equal $y-$ coefficient. We can either make the equal $x$ coefficient of equation $(i)$ and equation $(ii)$ . It is done because both the equations have different $x$ and $y$ coefficients but for simplifying and further calculation we have to make one of the coefficients equal. Either $x$ or $y$ for both the equations.
The calculated solutions can be verified to check whether it satisfies the equation or not which is as follows.
Put $x=2$ and $y=-1$ in equation $(i)$
$3x+y=5$
$\Rightarrow 3\times 2+\left( -1 \right)=5$
$\Rightarrow 6-1=5$
$5=5$
Now, put $x=2$ and $y=-1$ in equation $(ii)$
$x-2y=4$
$\Rightarrow 2-2\left( -1 \right)=4$
$\Rightarrow2+2=4$
$4=4$
Hence from the above calculation, the solutions are satisfied by both the equations, therefore the solutions calculated are correct.