# AP SSC Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variable

## Pair of Linear Equations in Two Variable

Two linear equations that have the same two variables are known as a pair of linear equations in two variables. Learn about Linear equations from AP SSC Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variable.

$a_{1}x+b_{1}x+c_{1}=0\,(a_{1}^2+b_{1}^2\neq 0)$ $a_{2}x+b_{2}x+c_{2}=0\,(a_{2}^2+b_{2}^2\neq 0)$

where $a_1,a_2,b_1,b_2,c_1,c_2$ are all real numbers.

We use the following methods to find solutions to a pair of linear equations:

• Model Method
• Graphical Method
• Algebraic methods – Substitution method and Elimination method

There exists a relation between the coefficients and nature of the system of equations. Following are the relationship:

• $\frac{a_1}{a_2}\neq \frac{b_1}{b_2}$, then the pair of linear equations is consistent
• $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$, then the pair of linear equations is inconsistent.
• $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, then the pair of linear equations is dependent and consistent.

Let us look at a few solved questions from the chapter to better understand a pair of linear equations.

### Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variable Solutions

1. Solve the following pair of equations by reducing them to a pair of linear equations.
$\frac{5}{x-1}+\frac{1}{y-2}=2$ $\frac{6}{x-1}+\frac{3}{y-2}=1$

Solution:

Let us consider the following

$\frac{1}{x-1}=u$ ……….(1)

$\frac{1}{y-2}=v$………….(2)

Hence, the equations becomes

$5u+v=2$…………..(3)

$6u+3v=1$…………(4)

From (1),

$v=2-5u$

Substituting the above in (4), we get

$6u-3(2-5u)=1$

Solving,

$6u-6+15u=1$ $21u-6=1$ $21u=7$ $u=\frac{7}{21}=\frac{1}{3}$

To find the value of v, substitute the above value of u in equation (3)

$5(\frac{1}{3})+v=2$ $\frac{5}{3}+v=2$ $v=2-\frac{5}{3}$ $v=\frac{6-5}{3}=\frac{1}{3}$

Hence,

$v=\frac{1}{3}$

To find the value of x and y, substitute the values of u and v in (1) and (2),

$u=\frac{1}{x-1}$ $\frac{1}{3}=\frac{1}{x-1}$ $x-1=3$ $x=4$ $v=\frac{1}{y-2}$ $\frac{1}{3}=\frac{1}{y-2}$ ${y-2}=3$ $y=5$

The value of x and y are 4 and 5 respectively for the given pair of equations.

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