Solving Linear equations in Two variables by elimination

A method of solving the linear equations in two variables by eliminating a variable from them is called the method of solving the linear equations in two variables by the elimination.

Introduction

The variables of two linear equations can be evaluated mathematically by the elimination. In this method, the coefficients of any one of the variables of both simultaneous linear equations are made same by multiplying them with suitable numbers. It helps us to eliminate the same terms from them and forms a linear equation in one variable.

Procedure

In elimination method, the linear equations in two variables are solved in three simple steps.

1. Multiply the two linear equations by the suitable numbers for making the coefficients of any one of the variables of both equations are same.
2. Add or subtract the linear equations to eliminate the same terms and it forms a linear equation in one variable. Then, simplify the linear equation in one variable to evaluate the variable.
3. Substitute the value of the variable in any one of the two given linear equations to get the value of the remaining variable.

Thus, the linear equations in two variables are solved by the elimination method mathematically.

Example

$x+3y-5 = 0$ and $7x-8y-6 = 0$ are two simultaneous linear equations in two variables $x$ and $y$. Shift the constants to right hand side of the equations.

1. $x+3y = 5$
2. $7x-8y = 6$

Now, let’s start solving the linear equations in two variables by elimination method.

Step – 1

Observe the coefficients of $x$ in both linear equations.

1. The coefficient of $x$ in $x+3y = 5$ is $1$.
2. The coefficient of $x$ in $7x-8y = 6$ is $7$.

We have to make the coefficients of $x$ in both linear equations should be same. If the equation $x+3y = 5$ is multiplied by $7$ and the equation $7x-8y = 6$ is multiplied by $1$, then the coefficients of $x$ in both linear equations become same.

$7 \times (x+3y) = 7 \times 5$

$\implies$ $7 \times x+ 7 \times 3y = 7 \times 5$

$\,\,\, \therefore \,\,\,\,\,\,$ $7x+21y = 35$

$1 \times (7x-8y) = 1 \times 6$

$\implies$ $1 \times 7x +1 \times (-8y) = 1 \times 6$

$\,\,\, \therefore \,\,\,\,\,\,$ $7x-8y = 6$

Step – 2

We have linear equations $7x+21y = 35$ and $7x-8y = 6$. If they are added, no terms are eliminated but if they are subtracted then the first terms are eliminated. So, subtract the equation $7x-8y = 6$ from $7x+21y = 35$.

$\implies$ $(7x+21y)-(7x-8y) = 35-6$

$\implies$ $7x+21y-7x+8y = 29$

$\implies$ $7x-7x+21y+8y = 29$

$\implies$ $\require{cancel} \cancel{7x}-\cancel{7x}+29y = 29$

$\implies$ $29y = 29$

It is a linear equation in one variable and solve it to find the value of $y$.

$\implies$ $y = \dfrac{29}{29}$

$\implies$ $\require{cancel} y = \dfrac{\cancel{29}}{\cancel{29}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $y = 1$

Step – 3

Substitute $y = 1$ in any one of the given two linear equations to find the value of $x$. In this case, $x+3y = 5$ is considered to replace the value of $y$ for finding the $x$ value.

$\implies$ $x+3(1) = 5$

$\implies$ $x+3 \times 1 = 5$

$\implies$ $x+3 = 5$

$\implies$ $x = 5-3$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 2$

Therefore, it is solved that $x = 2$ and $y = 1$. In this way, the linear equations in two variables are solved by elimination.

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