**Pair of Linear Equations in two variables**

**An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y. (We often denote the condition a and b are not both zero by a**^{2}**+ b**^{2}** ≠ 0)**

**For example, let us substitute x = 2 and y = 2 in the left hand side (LHS) of the equation 3x + 4y = 14**

**Then**

**LHS = 3(2) + 4(2) **

**= 6 + 8 **

**= 14,**

**which is equal to the right hand side (RHS) of the equation.**

**∴ x = 2 and y = 2 is a solution of the equation 3x + 4y = 14**

**Now let us substitute x = 1 and y = 1 in the equation 3x + 4y = 14. **

**Then,**

**LHS = 3(1) + 4(1) **

**= 3 + 4 **

**= 7**

**which is not equal to the RHS**

**∴ x = 1 and y = 1 is not a solution of the equation.**

**Geometrically, what does this mean? It means that the point (2, 2) lies on the line representing the equation 3x + 4y = 14, and the point (1, 1) does not lie on it. So, every solution of the equation is a point on the line representing it.**

**In fact, this is true for any linear equation, that is, each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.**

**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,**

**where a**_{1}**, b**_{1}**, c**_{1}**, a**_{2}**, b**_{2}**, c**_{2}** are all real numbers and**

**a**_{1}^{2}** + b**_{1}^{2}** ≠ 0, **

**a**_{2}^{2}** + b**_{2}^{2}** ≠ 0.**

**Example 5x + 34y – 7 = 0 and 5x – 22y + 9 = 0**

**To summarise the behaviour of lines representing a pair of linear equations in two variables and the existence of solutions**

**(i) x – y = 1 and x + y = 3 (The lines intersect) **

**(ii) the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).**

**x + y = 2 and x + y = 3 **

**(iii) the lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].**

**x + y = 2 and 2x + 2y = 4 **

**Lines represented by**

**a**_{1}**x + b**_{1}**y + c**_{1}** = 0 and**

**a**_{2}**x + b**_{2}**y + c**_{2}** = 0,**

**(i) intersecting, then**

**a**_{1}**/a**_{2}** ≠ b**_{1}**/b**_{2}

**(ii) coincident, then**

**a**_{1}**/a**_{2}** = b**_{1}**/b**_{2}** = c**_{1}**/c**_{2}

**(iii) parallel, then**

**a**_{1}**/a**_{2}** = b**_{1}**/b**_{2}** ≠ c**_{1}**/c**_{2}

**Substitution Method : We shall explain the method of substitution by taking some examples.**

**Solve the following pair of equations by substitution method:**

**x – y = 1 ⇒ (1)**

**x + y = 3 ⇒(2)**

**Solution :**

**Step 1 : **

**We pick either of the equations and write one variable in terms of the other.**

**Let us consider the Equation (2)**

**x + y = 3**

**and write it as x = 3 – y**

**Step 2 : **

**Substitute the value of x in Equation (1)**

**We get**

**(3 – y) – y = 1**

**3 – 2y = 1**

**2y = 2**

**∴ y = 1**

**Step 3 : **

**Substituting this value of y in Equation (1), we get**

**x = 2**

**Real Life application:**

**You have a specific area of plot where you are building your home. At one corner you plan to have experimental room. You don't know the dimension of it, but there is a relation where in area of a room gets reduced by 9 square units when length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the experimental room**

**Elimination Method**

**2x – y = 3 ⇒ (1)**

**x + y = 3 ⇒(2)**

**Step 1 : **

**Multiply Equation (1) by 1 and Equation (2) by 2 to make the coefficients of x equal. Then we get the equations:**

**2x – y = 3 ⇒(3)**

**2x + 2y = 6 ⇒(4)**

**Step 2 : **

**Subtract Equation (3) from Equation (4) to eliminate x, because the coefficients of y are the same. So, we get**

**3y = 3**

**y = 1**

**Step 3 : **

**Substituting this value of y in (1), we get**

**x = 2**

**Cross multiplication**

**Lines represented by**

**a**_{1}**x + b**_{1}**y + c**_{1}** = 0 and**

**a**_{2}**x + b**_{2}**y + c**_{2}** = 0**

**We can express it as**

**x/(b**_{1}**c**_{2}** - b**_{2}**c**_{1}**)**

**= y/(c**_{1}**a**_{2}** - c**_{2}**a**_{1}**)**

**= 1/(a**_{1}**b**_{2}** - a**_{2}**b**_{1}**) **

**Real Life application:**

**Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?**

**Let the speed of 1st car and 2nd car be u and v**

**Respective speed of both cars while they are travelling in same direction = (u - v) km/h**

**Respective speed of both cars while they are travelling towards each other = (u + v) km/h**

**Given: When the travel in same direction they meet after 5 hours**

**but require 1 hour when they travel towards each other.**

**Place is 100km apart**

**5(u - v) = 100**

**u - v = 20 km/h ⇒ (i)**

**1(u + v) = 100 ⇒ (ii)**

**Adding both the equations, we get**

**2u = 120 **

**u = 60 km/h ⇒ (iii)**

**Substituting this value in equation (ii), we obtain**

**v = 40 km/h**

**∴ speed of one car = 60 km/h and speed of other car = 40 km/h**