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Linear Equation

Explanation o...

Question

Explanation of the elimination method of linear equation in 2 variable.

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Solution

Step 1: Standard form of linear equation in two variables
Linear equation in two variables $x$ and $y$ is of the form
$a x + b y = c$
where $a, b, c$ are constant.

Step 2: Explanation of the elimination method
Suppose we have two simultaneous linear equations in two variables.
$\begin{array}{rcl} a_{1} x + b_{1} y & = & c_{1} - - - - - (1) \\ a_{2} x + b_{2} y & = & c_{2} - - - - - (2) \end{array}$
We have to eliminate one variable, either $x$ or $y$ from the two equations.
Suppose we eliminate $x$ . Then from the resulting equation we have to find the value of $y$ .
Then substitute the value of $y$ in any one of the equation $(1)$ or $(2)$ and solve the equation to find the value of $x$ .
Step 3: Elimination of $x$
Let us eliminate $x$ .
Multiply equation $(1)$ by $a_{2}$ . We get,
$\begin{array}{rcl} a_{2} (a_{1} x + b_{1} y) & = & a_{2} c_{1} \\ o r, a_{2} a_{1} x + a_{2} b_{1} y & = & a_{2} c_{1} - - - - - - (3) \end{array}$
Multiply equation $(2)$ by $a_{1}$ . We get,
$\begin{array}{rcl} a_{1} (a_{2} x + b_{2} y) & = & a_{1} c_{2} \\ o r, a_{1} a_{2} x + a_{1} b_{2} y & = & a_{1} c_{2} - - - - - (4) \end{array}$
Now, subtract equation $(3)$ from equation $(4)$
$\begin{array}{rcl} \frac{a_{1} a_{2} x + a_{1} b_{2} y = a_{1} c_{2} a_{2} a_{1} x + a_{2} b_{1} y = a_{2} c_{1}}{a_{1} b_{2} y - a_{2} b_{1} y = a_{1} c_{2} - a_{2} c_{1}} \end{array}$
$a_{1} a_{2} x$ and $a_{2} a_{1} x$ gets canceled out and we get the equation,
$a_{1} b_{2} y - a_{2} b_{1} y = a_{1} c_{2} - a_{2} c_{1}$

Step 4: Find out the value of $y$
$\begin{array}{rcl} a_{1} b_{2} y - a_{2} b_{1} y & = & a_{1} c_{2} - a_{2} c_{1} \\ o r, (a_{1} b_{2} - a_{2} b_{1}) y & = & a_{1} c_{2} - a_{2} c_{1} \\ o r, y & = & \frac{a_{1} c_{2} - a_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}} (D i v i d i n g b o t h s i d e b y a_{1} b_{2} - a_{2} b_{1}) \end{array}$

Step 5: Find out the value of $x$
Thus eliminating $x$ we get the value of $y$ .
Now we will put the value of $y$ in any one of the equation $(1)$ or $(2)$ and solve to get the value of $x$ .
Put the value of $y$ in equation $(1)$
$\begin{array}{rcl} a_{1} x + b_{1} (\frac{a_{1} c_{2} - a_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}) & = & c_{1} \\ o r, a_{1} x & = & c_{1} - b_{1} (\frac{a_{1} c_{2} - a_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}) \\ o r, a_{1} x & = & \frac{c_{1} (a_{1} b_{2} - a_{2} b_{1}) - b_{1} (a_{1} c_{2} - a_{2} c_{1})}{a_{1} b_{2} - a_{2} b_{1}} \\ o r, a_{1} x & = & \frac{(c_{1} a_{1} b_{2} - c_{1} a_{2} b_{1}) - (b_{1} a_{1} c_{2} - b_{1} a_{2} c_{1})}{a_{1} b_{2} - a_{2} b_{1}} \\ o r, a_{1} x & = & \frac{c_{1} a_{1} b_{2} - b_{1} a_{1} c_{2}}{a_{1} b_{2} - a_{2} b_{1}} (c_{1} a_{2} b_{1} and b_{1} a_{2} c_{1} gets cancelled out) \\ o r, a_{1} x & = & \frac{a_{1} (c_{1} b_{2} - b_{1} c_{2})}{a_{1} b_{2} - a_{2} b_{1}} \\ o r, x & = & \frac{c_{1} b_{2} - b_{1} c_{2}}{a_{1} b_{2} - a_{2} b_{1}} (Cancelling out a_{1} from both side) \end{array}$
Step 6: Conclusion
By elimination method $x = \frac{c_{1} b_{2} - b_{1} c_{2}}{a_{1} b_{2} - a_{2} b_{1}}$
$y = \frac{a_{1} c_{2} - a_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$
Hence, elimination method of linear equation in 2 variables is explained.

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