Question

How do you find the solution of system of equation $x+2y=12$ and $x-2y=0?$

Answer 1

Hint: As in above equation, for solving the value of $''x''$ and $''y''$ equalise the coefficient of any of the given variables, either equalise the coefficient of variable $''x'',$ like as in first equation the coefficient of $x$ is $1$ while in second equation the coefficient of x is also $2,$ so we don’t need to equalise them, so, we will subtract the later equation from the first equation, then by using BODMAS, determine the value of $''x''$ then by putting the value of $''x''$ in any of the equation determine the value of $''y''$

Complete step by step solution:
As per data given in the question,
As we have,
$x+2y=12...(1)$
$x-2y=0....(2)$
Here, we have to determine the value of $''x''$ and $''y''$
As, if we equalise the coefficient of $''x''$ in both equations,
We will get that,
The coefficient of $''x''$ in first equation is $1$ while the coefficient of $''x''$ in second equation is also$''1''$
So, there is no need to equalise them,
Now for getting the value of $''x''$ and $''y''$ we will subtract the later equation from the first equation,
Here,
We will get,
$\left[ x+2y \right]-\left[ x-2y \right]=12-0$
Here, left side elements will be subtracted from left side element while the right-side element will be subtracted from right side element,
So, subtracting them,
We will get,
$\left[ x+2y-x+2y \right]=12$
As, when we subtract the later equation from the first equation, then the sign of equation which is to be subtracted will be reversed,
So,
Here from above, $\left( +x \right)$ and $\left( -x \right)$ get cancelled,
Hence, we will get,
$4y=12$
As here the value of $4y$ is $12$
So, in order to determine the value of $''y''$ we need to shift the “4” which is in multiplication with $''y''$ in left side to right side, after shifting the 2 will come in division.
So,
$4y=12$
So, value of $''y''$ will be $=\dfrac{12}{4}=3$
Hence, after solving the above equation, we get value of $''y''$ as $3$
Now, for determining the value of $''x''$ we need to put the obtained value of $''y''$ in any of the given equation,
Putting the value of $''y''$ in first equation,
We will get,
$\left( x+2y \right)=12$
$\Rightarrow \left( x+2\times 3 \right)=12$
$\Rightarrow x+6=12$
Now separating the like terms,
We will get,
$x=12-6$
So, value of “x” will be $=6$

Hence, value of $''x''$ and $''y''$ in above equation is $6$ and $3$ respectively.

Note: If we have an equation,
$\Rightarrow 2x+3y=4$
If we multiply all terms of the equation by any of the numbers then the value of the equation will not change.
Like,
If we multiply all terms by $2$
We will get,
$2\times \left[ 2x+3y \right]=4\times 2$
$\Rightarrow 4x+6y=8$
But always remember that,
While multiplying the equation by any of the constants make sure that you are multiplying each and every terms of the equation.
When we subtract any equation from the other equation, then the sign of the equation which is subtracting from another equation gets reversed.
Like, if we have to subtract $x+y=2$ from $2x+4y=5$
Then while subtracting,
Sign of first equation get reversed,
So, we will get,
$2x+4y-x-y=5-2$
Hence, positive sign is replaced by negative sign, and vice versa.