 Introduction to Pair of Linear Equations in Two Variables

Linear equations in two variables are equations which can be expressed as ax + by + c = 0, where a, b and c are real numbers and both a, and b are not zero. The solution of such equations is a pair of values for x and y  which makes both sides of the equation equal.

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Let’s look at the solutions of some linear equations in two variables. Consider the equation 2x + 3y = 5. There are two variables in this equation, x and y.

• Scenario 1: Let’s substitute x = 1 and y = 1 in the Left Hand Side (LHS) of the equation. Hence, 2(1) + 3(1) = 2 + 3 = 5 = RHS (Right Hand Side). Hence, we can conclude that x = 1 and y = 1 is a solution of the equation 2x + 3y = 5. Therefore, x = 1 and y = 1 is a solution of the equation 2x + 3y = 5.
• Scenario 2: Let’s substitute x = 1 and y = 7 in the LHS of the equation. Hence, 2(1) + 3(7) = 2 + 21 = 23 ≠ RHS. Therefore, x = 1 and y = 7 is not a solution of the equation 2x + 3y = 5.

Geometrically, this means that the point (1, 1) lies on the line representing the equation 2x + 3y = 5. Also, the point (1, 7) does not lie on this line. In simple words, every solution of the equation is a point on the line representing it.

To generalize, 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.

Pair of Linear Equations in Two Variables

Here is a situation: The number of times Ram eats a mango is half the number of rides he eats an apple. He goes to the market and spends Rs. 20. If one mango costs Rs.3 and one apple costs Rs.4, then how many mangoes and apples did Ram eat?

Let’s say that the number of apples that Ram ate is y and the number of mangoes is x. Now, the situation can be represented as follows:

y = (½)x … {since he ate mangoes (x) which were half the number of apples (y)}
3x + 4y = 20 … {since each apple (y) costs Rs.4 and mango (x) costs Rs.3}

Both these equation together represent the information about the situation. Also, these two linear equations are in the same variables, x and y. These are known as a ‘Pair of Linear Equations in Two Variables’.

To generalize them, a pair of linear equations in two variables x and y is:

a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0.

Where a1, b1, c1, a2, b2, c2 are all real numbers and a12+ b12 ≠ 0, a22+ b22 ≠ 0. Some examples of a pair of linear equations in two variables are:

• 2x + 3y – 7 = 0 and 9x – 2y + 8 = 0
• 5x = y and –7x + 2y + 3 = 0

Geometric Representation of a Pair of Linear Equations in Two Variables

By now, we know that the geometrical or graphical representation of linear equations in two variables is a straight line. Hence, a pair of linear equations in two variables will be two straight lines which are considered together. We also know that when there are two lines in a plane:

• The two lines will intersect at one point. {Fig.1 (a)}
• They will not intersect, i.e., they are parallel. {Fig.1 (b)}
• The two lines will be coincident. {Fig.1 (c)} Fig. 1

Linear equations can be represented both algebraically and geometrically. Let’s see how. Going back to our earlier example of Ram, let’s try to represent the situation both algebraically and geometrically.

SolutionAlgebraic Representation

The pair of equations formed is: y = (1/2)x
So, 2y = x
Hence, x – 2y = 0 … (1)
3x + 4y = 20 … (2)

Geometric Representation

To represent these equations graphically we need at least two solutions for each equation. These solutions are listed in the tables below:

 x 0 2 y = (1/2)x 0 1

 x 0 4 y = (20 – 3x)/4 5 2

Now, we take a graph paper and plot the points A(0, 0), B(2, 1) and P(0, 5), Q(4, 2), corresponding to the solutions in tables above. Next, we draw the lines AB and PQ as shown below. Fig. 2

This is the geometric representation of the equations x – 2y = 0 and 3x + 4y = 20.

Solved Examples for You

Question: Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically.

Solution: Let Aftab’s present age be x and his daughter’s present age by y.

7 years ago, Aftab’s age = x – 7
His daughter’s age = y – 7

Also, according to the question,
(x – 7) = 7(y – 7)
So, x – 7 = 7y – 49
Hence, x – 7y = 42

Now, three years later,
Aftab’s age = x + 3
His daughter’s age = y + 3

Also, according to the question,
(x + 3) = 3(y + 3)
So, x + 3 = 3y + 9
Hence, x – 3y = 6

Therefore, the situation can be algebraically represented as:

• x – 7y = 42
• x – 3y = 6
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