Pair of Linear Equations in Two Variables

Introduction and NCERT Exercise 3.1

An expression in the form of `ax+by+c=0` where `a, b` and `c` are real numbers, and `a` and `b` are not both zero (`a^2+b^2!=0`), is called a LINEAR EQUATION IN TWO VARIABLES.

Example:

`2x+3y-5=0`

Here, `a=2, b=3` and `c =-5` and are real numbers.

And `2^2+3^2!=0`

Solution of the above linear equation `2x+3y-5=0`

Let, substitute `x=1` and `y=1` in the linear equation `2x+3y-5=0`

Therefore,

`2xx1+3xx1-5=0`

`=>2+3+5=0`

`=>0=0`

This means that LHS = RHS.

Thus, `x=1` and `y=1` is the solution of given equation.

Geometrically meaning of Linear Equation

Geometrical meaning of the given, equation, `2x+3y-5=0` is point (1, 1) lies one the line representing the equation `2x+3y-5=0`.

So, every solution of the equation is a point on the line representing it.

Thus, Each solution (`x, y`) of linear equation in two variables, `ax+by+c=0` corresponds to a point on the line representing the equation, and vice versa.

A Pair of Linear Equations in Two Variables

The two linear equations are in the same two variables `x` and `y` are called A Pair of Linear Equations in Two Variables.

The general form of a pair of linear equations in two variables `x` and `y` is

`a_1x+b_1y+c_1=0`

And `a_2x+b_2y+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:

`2x+3y-7=0` and `9x-2y+8=0`

How does a pair of linear equation in two variables look like Geometrically?

When a pair of linear equation in two variables is represented geometrically, only one of the following three possibilities can happen:

(i) The two lines will intersect at one point.

(ii) The two lines will not intersect, i.e. they are parallel.

(iii) The two lines will be coincident.

NCERT Exercise 3.1

Question: 1. 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 and graphically.

Solution:

Let, present age of Aftab `=x`

And present age of his daughter `=y`

Therefore,

Seven years ago age of Aftab `=x-7`

And seven years ago age of his daughter `=y-7`

As given in question,

`(x-7)=7(y-7)`

`=> x-7 = 7y - 49`

`=>x-7y = -49+7 `

`=>x-7y = -42` --------(i)

Three years from now,

Age of Aftab `=x+3`

And age of his daughter `=y+3`

Again, as given in question,

`(x+3) = 3(y+3)`

`=>x+3 = 3y + 9`

`=> x-3y = 9-3`

`=>x+3y = 6` -------(ii)

Thus, algebraic representation of given situation is

`x-7y = -42` And

`x-3y = 6`

Now,

For, `x-7y = -42`

`=> x = -42+7y` -------(iii)

If we take the value of y = 5, 6 and 7

Then table of solution for this equation (iii).

`x`	`-7`	0	7
`y`	5	6	7

For, `x-3y = 6`

`=>x = 6+3y`

If we take the value of y = 0, -1 and -2

Then table of solution for above equation is

`x`	6	3	0
`y`	0	`-1`	`-2`

Thus, algebraic representation of situation

`x-7y = -42` and

`x-3y = 6`

Graphical representation of equations:

Question: 2. The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.

Solution:

Let, the cost of a bat = Rs `x`.

And let the cost of a ball = Rs `y.

Thus, as per question:

`3x+6y = 3900.

`=> 3(x+2y) = 3900`

`=>x+2y = 3900/3`

`=> x+2y = 1300` ----------(i)

And, `x+3y = 1300` --------(ii)

Now, from equation (i)

`x = 1300-2y`

Thus table of solution for above equation (i)

`x`	6	3	0
`y`	0	`-1`	`-2`

And from equation (ii)

`x = 1300 - 3y`

Thus table of solution for above equation (ii)

`x`	400	100	–200
`y`	300	400	500

Algebraic representation of situation:

`3x+6y = 3900` or `x+2y = 1300`

And, `x+3y = 1300`

Graphical representation of equations:

Question: 3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Solution:

Let the cost of 1 kg of apples = Rs `x`

And cost of cost of 1 kg of grapes = Rs `y`

Therefore, according to question

`2x + y = 160` --------(i)

And after one month

`4x + 2y = 300` -------(ii)

`=> 2(2x+y) = 300`

`=> 2x + y = 300/2`

`=> 2x + y = 150` --------(iii)

Now, from eqution (i)

`x = (160 ? y)/2`

Thus table of solution for this equation

`x`	50	60	70
`y`	60	40	20

Now, from equation (ii)

`4x+2y = 300`

`=> x = (300-2y)/4` ---(iv)

Table of solution for equation

`x`	50	60	70
`y`	50	30	10

Thus, algebraic representation of the given situations

`2x+y = 160` and

`4x + 2y = 300` or `2x + y = 150`

Geometrical representation of the given situation

Graphical Method of Solution of a Pair of Linear Equations

There are three types of a pair of linear equations,

(i) Consistent pair of Linear Equations

(ii) In consistent pair of Linear Equations

(iii) Dependent pair of Linear Equations

(i) Consistent Pair of Linear Equations

A pair of linear equations in two variables which has one and only one solution, i.e. has a unique solution is called CONSISTENT PAIR OF LINEAR EQUATIONS IN TWO VARIABLES.

When a graph is plot for the consistent pair of linear equations, line intersects in a single point and thus has a unique solution.

Example:

`x-2y = 0` and `3x+4y =0`

This pair of linear equation has a unique solution (4, 2).

Here, `a_1 = 1, a_2 = 3, b_1 =-2` and `b_2 = -20`

[`a_1, a_2, b_1` and `b_2` are the coefficients of equations]

In this case, `a_1/a_2 != b_1/b_2`

i.e. `1/3 != (-20)/4`

(ii) In consistent pair of Linear Equations

A pair of linear equations in two variables which has no solution, is called an INCONSISTENT PAIR OF LINEAR EQUATIONS.

When a graph of an Inconsistent pair of Linear equations is plotted, the lines so formed are parallel.

Example:

`x+2y-4=0` and

`2x+4y-12=0`

This pair of linear equations has no solution.

Comparing this pair of linear equations with general form of equations, `a_1x+b_1x+c_1=0` and `a_2+b_2x+c_2=0`, we get

`a_1 = 1, b_1 = 2`, and `c_1=-4` and `a_2= 2, b_2 = 4`, and `c_2 = -12`

Thus, when `a_1/a_2=b_2/b_2!=c_1/c_2`

i.e. `1/2=2/4!=(-4)/(-12)`

Thus, when `a_1/a_2=b_2/b_2!=c_1/c_2`, the pair of linear equations in two variables has no solution, and lines come parallel when a graph is plotted. And such pair of linear equations in two variables is called inconstant pair of equations.

(iii) Dependent pair of Linear Equations

A pair of linear equations in two variables which has infinitely many distinct common solutions, is called DEPENDENT PAIR OF LINEAR EQUATIONS IN TWO VARAIBALES.

When a graph plotted for a dependent pair of linear equations, coincident lines are formed.

Example:

`2x+3y-9 =0` and `4x+6y-18=0`

Comparing this pair of linear equations with general form of equations, `a_1x+b_1x+c_1=0` and `a_2+b_2x+c_2=0`, we get

`a_1 = 2, b_1=3, c_1 = -9`

and `a_2= 4, b_2 =6, c_2 = -18`

Here,

`a_1/a_2 = b_1/b_2=c_1/c_2`

i.e.` 2/4 = 3/6 = (-9)/(-18)`

Thus, when `a_1/a_2 = b_1/b_2=c_1/c_2`

the pair of linear equations in two variables has infinitely many solutions and lines coincident when graph is plotted.

If a pair of linear equations in two variables be

`a_1x+b_1y+c_1 =0` and

`a_2x+b_2y+c_2=0`, then

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