CBSE Class 10 Mathematics Chapter 3 - Algebraic methods of solving a pair of linear equations - Notes

Algebraic Methods of Solving a Pair of Linear Equations:

• The graphical method become inconvenient when the point representing the solution of the linear equations has non-integral coordinates like (√3,2√7), (—1.75, 3.3) or fraction (4/13, 1/19). Reading and interpreting such coordinates can lead to errors.
• Due to the inconvenience of non-integral coordinates and the possibility of mistakes, alternative methods are required. Algebraic methods provide a systematic way to find solutions to linear equations, especially useful when working with equations with non-integral or complex solutions.

Substitution Method:

Step 1 - Express One Variable in Terms of the Other:

• The first step in the substitution method is to find the value of one variable (let's say y) in terms of the other variable (x).
• You can choose either of the two equations to do this, depending on which is more convenient.

Step 2 - Substitute and Reduce:

• After find the expression for one variable (y in terms of x), substitute this expression into the other equation. This will create an equation in a single variable (x) that can be solved.
• In some cases, this substitution may lead to statements with no variables. If such a statement is true, it implies that the pair of linear equations has infinitely many solutions. If the statement is false, it indicates that the pair of linear equations is inconsistent.

Step 3 - Solve for the Other Variable:

• Once you have solved the equation in terms of x (Step 2), you can substitute the value of x back into the equation used in Step 1 to find the value of the other variable, y.
• Because in this process substituting the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. So, it is known as Substitution Method. This systematic process simplifies the solution of the equations.

Example 1: Solve the following pair of equations by substitution method:

7x – 15y = 2 ……………….. (i)

x + 2y = 3  ……………………(ii)

Solution: Take equation (ii) and solve for variable x,

x + 2y = 3

x = 3 – 2y ……………….. (iii)

Step 2: substitute expression (iii) from equation (i)

7(3 – 2y) – 15y = 2

21 – 14y – 15y = 2

– 14y – 15y = 2 – 21

– 29y  =  – 19

Step 3: Find the value of the first variable (x) using the obtained value (y). Put y value in equation (iii)

Example 2: 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.” Represent this situation algebraically and graphically by the method of substitution.

Solution: Let Aftab present age in year is a and his daughter present age in year d.

According to question first condition:

7 year before age for Aftab is a — 7 and his daughter age d — 7

a — 7 = 7(d — 7)              (because Aftab seven time old)

Þ a — 7 = 7d — 49

Þ a —7d =  — 42 …………………. (i)

According to question second condition:

3 year after age for aftab is a + 3 and his daughter age d + 3

a + 3 = 3(d + 3)              (because Aftab three time old)

Þ a + 3 = 3d + 9

Þ a — 3d = 6 ……………… (ii)

Þ a = 3d + 6 ………………… (iv)

Putting a value from equation (iv) to equation (i)

(3d + 6) —7d =  — 42

Þ 3d + 6 —7d =  — 42

Þ 3d—7d =  — 42  — 6

Þ —4d =  — 48

Þ d =  12   (Daughter age)

Putting d value in equation (ii)

Þ a — 3(12) = 6

Þ a — 36 = 6

Þ a = 6+36 = 42     (Aftab Age)

Graphically representation:

Table of value for equation (i) a —7d =  — 42

a	-7	0	7
d	5	6	7

Table of value for equation (ii) a — 3d = 6

a	0	3	6
d	-2	-1	0

Plot graph:

Example 3: In a shop the cost of 2 pencils and 3 erasers is Rupees 9 and the cost of 4 pencils and 6 erasers is Rupees 18. Find the cost of each pencil and each eraser.

Solution: Let each pencil cost is x and each eraser cost is y.

According to question first condition:   2x + 3y = 9  ………………… (i)

According to question Second condition:   4x + 6y = 18  ………………… (ii)

From equation (i) 

Substitute x value from equation (ii)

We cannot obtain a specific value for x because both given equation are same. Therefore equation (i) and (ii) have infinitely many solutions.

Note: We can solve this question by using ratio of coefficient for coincident lines.

In equation (i) and (ii) Ratio of coefficient are same. This ratio shows that linear equation has infinitely many solutions. Hence, infinitely many cost of pencil and eraser is possible.

Example 4: Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails cross each other?

Solution: Both rails are cross each other if given equation has a unique solution.

Given equation are:

x + 2y – 4 = 0 ……………………. (i)

2x + 4y – 12 = 0 ………………… (ii)

From equation (i)

x + 2y – 4 = 0

x = 4 – 2y

Substitute the value from equation (ii)

 2(4 – 2y) + 4y – 12 = 0

Þ 8 – 4y + 4y = 12

Þ 8 – 12  = 4y – 4y     (by combine like terms)

Þ  – 4  = 0

 – 4  does not equal to 0, so there is no solution for y. so two rails will not cross each other.