Pair of Linear Equations in Two Variables Class 10 Notes CBSE Maths Chapter 3 (Free PDF Download)

Linear Equation

• A linear equation in two variables is defined as an equation with the form ${\text{ax}} + {\text{by}} + {\text{c}} = 0$, where ${\text{a}},{\text{b}}$ and ${\text{c}}$ are real numbers and both a and ${\text{b}}$ are nonzero.

Solution of an Equation

• Each two-variable solution ( ${\text{x}},{\text{y}}$ ) of a linear equation. A point on the line expressing the equation corresponds to $ax + by + c = 0$ and vice-versa. 

Pair of Linear Equations in Two Variables

• The general form for a pair of linear equations in two variables $x$ and $y$ is ${a_1}x + {b_1}y + {c_1} = 0$ And ${a_2}x + {b_2}y + {c_2} = 0$

• From a geometric standpoint, they resemble the following:

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Graphical Method of Solutions

• $x - 2y = 0$

$3x + 4y = 20$

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The solution is $(4,2)$, the point of intersection.

• To summarise the behaviour of two-variable lines expressing a pair of linear equations:

• The lines might cross at a single place. The pair of equations has a unique solution in this situation (consistent pair of equations).

• There's a chance the lines aren't connected. The equations have an unlimited number of solutions (dependent (consistent) pair of equations) in this situation.

Substitution Method

The following are the steps:

Step 1: Find the value of one variable, say ${\text{y}}$ in terms of the other variable, i.e., ${\text{x}}$ from either equation, whichever is convenient.

Step 2: Substitute this value of ${\text{y}}$ in the other equation, and reduce it to an equation in one variable, i.e., in terms of ${\text{x}}$, which can be solved. You could come across sentences that don't have any variables. If this assertion is correct, the pair of linear equations have an unlimited number of solutions. The pair of linear equations is illogical if the assertion is incorrect. 

Step 3: Substitute the value of $x$ (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

For more understanding the concept, we are going to see the example to solve two equations $x - 2y = 8$ and $x + y = 5$ with the help of substitution method.

$x - 2y = 8$  equation (1)

$x + y = 5$   equation (2)

From equation (2), $x = 5 - y$

Substituting this value in equation (1),

$x - 2y = 8 $

$\Rightarrow 5 - y - 2y = 8$

$\Rightarrow 5 - 3y = 8 $

$\Rightarrow - 3y = 8 - 5$

$\Rightarrow y=  - 1$

Put this value in equation (2),

$x - 1 = 5$

$x = 5 + 1$

$x=6$

Hence $x=6$ and $y=-1$

Elimination Method

Steps in the elimination method:

Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either ${\text{x}}$ or ${\text{y}}$ ) numerically equal.

Step 2: Then subtract or add one equation from the other to eliminate one variable. Go to Step 3 if you receive an equation in one variable. If we get a true statement with no variables in Step 2, the original set of equations contains an unlimited number of solutions. 

If we have a false statement with no variable in Step 2, the original set of equations has no solution, which means it is inconsistent.

Step 3: Solve the equation in one variable ( ${\text{x}}$ or ${\text{y}}$ ) so obtained to get its value.

Step 4: Substitute this value of $x$ (or $y$ ) in either of the original equations to get the value of the other variable.

Solved Example: 

 $x - 2y = 8$ equation (1) 

 $2x + y = 5$ equation (2) 

Multiply equation (2) by 2

$  {(2x + y = 5) \times 2} $

$  {\Rightarrow 4x + 2y = 10}$

Add this equation to equation (1), we get,

$x - 2y = 8$

$4x + 2y = 10$

Now on solving the above equation, we get

$5x = 18$

$x = \dfrac{{18}}{5}$

Put this value in equation (2)

  $y = 5 - 2x$

  $\Rightarrow y = 5 - 2 \times \dfrac{{18}}{5} $

  $\Rightarrow  y = 5 - \dfrac{{36}}{5}$

$\Rightarrow y =  - \dfrac{{11}}{5}$

$\therefore x = \dfrac{{18}}{5}$ and $y =  - \dfrac{{11}}{5}$

Cross Multiplication Method

Steps:

Write the given equations in the form

${a_1}x + {b_1}y + {c_1} = 0$

And ${a_2}x + {b_2}y + {c_2} = 0$

Using the diagram as a guide

Cross Multiplication

Write Equations as

$\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$

Find $x$ and $y$, provided ${a_1}{b_2} - {a_2}{b_1} \ne 0$

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