Algebraic Methods of Solving a Pair of Linear Equations

  1. Substitution method

If we have a pair of Linear Equations with two variables x and y, then we have to follow these steps to solve them with the substitution method-

Step 1: We have to choose any one equation and find the value of one variable in terms of other variable i.e. y in terms of x.

Step 2: Then substitute the calculated value of y in terms of x in the other equation.

Step 3: Now solve this Linear Equation in terms of x as it is in one variable only i.e. x.

Step 4: Substitute the calculate value of x in the given equations and find the value of y.

EXAMPLE

y – 2x = 1

x + 2y = 12

y = 2x + 1.

Substitute y in the second equation to get x.

x + 2 × (2x + 1) = 12

⇒ 5x + 2 = 12

x + 2 = 12
⇒ x = 2

y = 2 \times 2 + 1

⇒y = 5

So, (2, 5) is the required solution of the pair of linear equations

  1. Elimination method

In this method, we solve the equations by eliminating any one of the variables.

Step 1: Multiply both the equations by such a number so that the coefficient of any one variable becomes equal.

Step 2: Now add or subtract the equations so that the one variable will get eliminated as the coefficients of one variable are same.

Step 3: Solve the equation in that leftover variable to find its value.

Step 4: Substitute the calculated value of variable in the given equations to find the value of the other variable.

Consider x + 2y = 8 and 2x – 3y = 2

Step 1: Make the coefficients of any variable the same by multiplying the equations with constants. Multiplying the first equation by 2, we get,

2x + 4y = 16

Step 2: Add or subtract the equations to eliminate one variable, giving a single variable equation.
Subtract second equation from the previous equation
\begin{array}{*{20}{c}}{2x}& + &{4y}& = &{16}\\{2x}& - &{3y}& = &2\\ - &{}& + &{}& - \\\hline{0x}& + &{7y}& = &{14}\end{array}
Step 3: Solve for one variable and substitute this in any equation to get the other variable.

y = 2,

x = 8 – 2 y

⇒ x = 8 – 4

⇒ x = 4

(4, 2) is the solution.

  1. Cross multiplication method

Given two equations in the form of

{a_1}x + {b_1}y + {c_1} = 0 and {a_2}x + {b_2}y + {c_2} = 0, where

x = \frac{{({b_1}{c_2} - {b_2}{c_1})}}{{({a_1}{b_2} - {a_2}{b_1})}},y = \frac{{({c_1}{a_2} - {c_2}{a_1})}}{{({a_1}{b_2} - {a_2}{b_1})}}

Comparing the ratios of coefficients of a Linear Equation

  1. Equations Reducible to a Pair of Linear Equations in Two Variables

There is some pair of equations which are not linear but can be reduced to the linear form by substitutions.

Given equations

\frac{a}{x} + \frac{b}{y} = c

We can convert these type of equations in the form of ax + by + c = 0

Let\frac{1}{x} = mand\frac{1}{y} = n

Now after substitution the equation will be

am + bn = c

It can be  solved by any of the method of solving Linear Equations

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