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Solving Linear equations in Two variables by substitution

A method of solving the linear equations in two variables by substituting the values of the variables is called the method of solving the linear equations in two variables by the substitution.

Introduction

The two variables of two simultaneous linear equations can be solved in mathematics by the substitution. In this method, the equivalent value of one variable is substituted to find the value of the second variable, and the evaluated value of variable is used to find the remaining variable by the substitution.

Procedure

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

  1. Take an equation from the given two linear equations, then express a variable in terms of the remaining variable.
  2. Substitute the value of the variable in the remaining linear equation for transforming the linear equation from two variables to one variable, and then solve the value of the variable.
  3. Substitute the value of the variable in any one of the given equations to find the value of the remaining variable.

In this way, the linear equations in two variables are solved mathematically by the substitution.

Example

$2x-5y-4 = 0$ and $3x-2y+16 = 0$ are two simultaneous linear equations in two variables $x$ and $y$ but they are unknown.

Step – 1

Consider any one of the given linear equations and express either $x$ or $y$ in terms of the remaining variable. In this case, $3x-2y+16 = 0$ is considered and the variable $x$ is expressed in terms of $y$.

$3x-2y+16 = 0$

$\implies$ $3x-2y = -16$

$\implies$ $3x = 2y-16$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = \dfrac{2y-16}{3}$

Step – 2

Substitute the value of $x$ in the remaining linear equation for converting the two variables linear equation into one variable linear equation.

$2x-5y-4 = 0$

$\implies$ $2\Bigg(\dfrac{2y-16}{3}\Bigg)-5y-4 = 0$

$\implies$ $\dfrac{2}{3}(2y-16)-5y-4 = 0$

$\implies$ $3 \times \Bigg(\dfrac{2}{3}(2y-16)-5y-4\Bigg) = 3 \times 0$

$\implies$ $3 \times \dfrac{2}{3}(2y-16)-3 \times 5y-3 \times 4 = 0$

$\implies$ $\dfrac{3 \times 2}{3}(2y-16)-15y-12 = 0$

$\implies$ $\require{cancel} \dfrac{\cancel{3} \times 2}{\cancel{3}}(2y-16)-15y-12 = 0$

$\implies$ $2(2y-16)-15y-12 = 0$

$\implies$ $2 \times 2y-2 \times 16-15y-12 = 0$

$\implies$ $4y-32-15y-12 = 0$

$\implies$ $4y-15y-32-12 = 0$

$\implies$ $-11y-44 = 0$

$\implies$ $-44 = 11y$

$\implies$ $11y = -44$

$\implies$ $y = \dfrac{-44}{11}$

$\implies$ $\require{cancel} y = \dfrac{\cancel{-44}}{\cancel{11}}$

$\,\,\, \therefore \,\,\,\,\,\, y = -4$

Step – 3

Now, substitute the value of the variable $y$ in any of the given linear equations to evaluate the variable $x$.

Substitute $y = -4$ in the given linear equation $2x-5y-4 = 0$

$\implies$ $2x-5(-4)-4 = 0$

$\implies$ $2x+20-4 = 0$

$\implies$ $2x+16 = 0$

$\implies$ $2x = -16$

$\implies$ $x = \dfrac{-16}{2}$

$\implies$ $\require{cancel} x = \dfrac{\cancel{-16}}{\cancel{2}}$

$\,\,\, \therefore \,\,\,\,\,\, x = -8$

Therefore, $x = -8$ and $y = -4$. In this way, the linear equations in two variables are solved by the substitution method.

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