Class 10 Substitution Method

Sanjay Kumawat November 4, 2026 10th Mathematics(English) No Comments

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1 1.Class 10 Substitution Method

1.1 2.Class 10 Substitution Method Examples:

1.2 3.Practice Problems of Class 10 Substitution Method for Students:

1.2.1 4.Frequently Asked Questions Related to Class 10 Substitution Method:

1.2.2 Q:1.What is the substitution method?

1.2.3 Q:2.Write the working rule of substitution method

1.2.4 Q:3.What is meant by algebraic methods?

1.2.5 5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

1.Class 10 Substitution Method

Class 10 Substitution Method

Learn,”Class 10 Substitution Method” Understand linear equations,pairs of linear equations,method of solving pair of linear examples easily.
Understand the “Class 10 Substitution Method” chapter 3.2 of Class 10 Mathematics in easy Hindi and English.Linear equations ,with examples of linear equations in two variables!
“Before moving to Part 2,read our basic introduction to ” Pair of Linear Equations in Two Variables 10th.”

Also Read This Article:- Substitution method

2.Class 10 Substitution Method Examples:

Example:1.Solve the following pair of linear equations by the substitution method.
Example: 1 (i). $x + y = 14, \quad x - y = 4$
Solution: $x + y = 14, \quad x - y = 4$
$x = 4 + y \quad \dots \text{(2)}$
Putting the value of $x$ in equation (1), we get:
$4 + y + y = 14 \implies 2y = 14 - 4$
$\implies y = \frac{10}{2} = 5$
Substitute this value of $y$ in equation (2), we get:
$x = 4 + 5 = 9$
$x = 9, \quad y = 5$
Example: 1 (ii). $s - t = 3, \quad \frac{s}{3} + \frac{t}{2} = 6$
Solution: $\frac{s}{3} + \frac{t}{2} = 6 \quad \dots \text{(1)}$
$s = t + 3 \quad \dots \text{(2)}$
Putting value of $s$ in equation (1), we get:
$\frac{t + 3}{3} + \frac{t}{2} = 6 \implies \frac{t}{3} + 1 + \frac{t}{2} = 6$
$\implies \frac{2t + 3t}{6} = 5 \implies t = \frac{30}{5} = 6$
Put the value of $t$ in equation (2), we get:
$s = 6 + 3 = 9$
$t = 6, \quad s = 9$
Example: 1 (iii). $3x - y = 3, \quad 9x - 3y = 9$
Solution: $3x - y - 3 = 0 \quad \dots \text{(1)}$
$9x - 3y - 9 = 0 \quad \dots \text{(2)}$
$a_1 = 3, \quad b_1 = -1, \quad c_1 = -3$
$a_2 = 9, \quad b_2 = -3, \quad c_2 = -9$
$\frac{a_1}{a_2} = \frac{3}{9} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-3}{-9} = \frac{1}{3}$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
Where x takes any value, i.e., infinitely many solutions.
Example: 1 (iv). $0.2x + 0.3y = 1.3$
$0.4x + 0.5y = 2.3$
Solution: $0.2x + 0.3y - 1.3 = 0$
$0.4x + 0.5y - 2.3 = 0$
$a_1 = 0.2, \quad b_1 = 0.3, \quad c_1 = -1.3$
$a_2 = 0.4, \quad b_2 = 0.5, \quad c_2 = -2.3$
$\frac{a_1}{a_2} = \frac{0.2}{0.4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{5}, \quad \frac{c_1}{c_2} = \frac{13}{23}$
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \quad \text{unique solution}$
$0.2x + 0.3y = 1.3 \quad \dots \text{(1)}$
$0.4x = 2.3 - 0.5y \implies x = \frac{2.3 - 0.5y}{0.4} \quad \dots \text{(2)}$
Putting the value of x in (1), we get:
$0.2\left(\frac{2.3 - 0.5y}{0.4}\right) + 0.3y = 1.3$
$\implies \frac{2.3 - 0.5y}{2} + 0.3y = 1.3$
$\implies 2.3 - 0.5y + 0.6y = 2.6$
$\implies 0.1y = 2.6 - 2.3$
$\implies 0.1y = 0.3$
$\implies y = \frac{0.3}{0.1} = 3$
Putting value of y in (2), we get:
$x = \frac{2.3 - 0.5 \times 3}{0.4} = \frac{2.3 - 1.5}{0.4} = \frac{0.8}{0.4} = 2$
Example: 1 (v). $\sqrt{2}x + \sqrt{3}y = 0$
$\sqrt{3}x - \sqrt{8}y = 0$
Solution: $\sqrt{2}x + \sqrt{3}y = 0 \quad \dots \text{(1)}$
$\sqrt{3}x - \sqrt{8}y = 0 \quad \dots \text{(2)}$
$a_1 = \sqrt{2}, \quad b_1 = \sqrt{3}, \quad c_1 = 0$
$a_2 = \sqrt{3}, \quad b_2 = -\sqrt{8}, \quad c_2 = 0$
$\frac{a_1}{a_2} = \frac{\sqrt{2}}{\sqrt{3}}, \quad \frac{b_1}{b_2} = \frac{\sqrt{3}}{-\sqrt{8}} \implies \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
Unique solution.
$\sqrt{3}x = \sqrt{8}y \implies x = \frac{\sqrt{8}}{\sqrt{3}}y \quad \dots \text{(3)}$
Put value of x in (1), we get:
$\sqrt{2}\left(\frac{\sqrt{8}}{\sqrt{3}}y\right) + \sqrt{3}y = 0$
$\implies \frac{\sqrt{16}}{\sqrt{3}}y + \sqrt{3}y = 0$
$\implies 4y + 3y = 0 \implies 7y = 0$
$\implies y = 0$
Putting value of y in (3), we get:
$x = \frac{\sqrt{8}}{\sqrt{3}}(0) \implies x = 0$
Example: 1 (vi). $\frac{3x}{2} - \frac{5y}{3} = -2$
$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
Solution: $\frac{3x}{2} - \frac{5y}{3} + 2 = 0 \quad \dots \text{(1)}$
$\frac{x}{3} + \frac{y}{2} - \frac{13}{6} = 0 \quad \dots \text{(2)}$
$a_1 = \frac{3}{2}, \quad b_1 = -\frac{5}{3}, \quad c_1 = 2$
$a_2 = \frac{1}{3}, \quad b_2 = \frac{1}{2}, \quad c_2 = -\frac{13}{6}$
$\frac{a_1}{a_2} = \frac{\frac{3}{2}}{\frac{1}{3}} = \frac{9}{2}, \quad \frac{b_1}{b_2} = \frac{-\frac{5}{3}}{\frac{1}{2}} = -\frac{5}{3} \times \frac{2}{1} = -\frac{10}{3}$
$\implies \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
Unique solution
From (1): $\frac{3x}{2} = -2 + \frac{5y}{3}$
$\implies x = \frac{2}{3}\left(-2 + \frac{5y}{3}\right)$
$\implies x = -\frac{4}{9} + \frac{10}{9}y \quad \dots \text{(3)}$
Putting the value of x in equation (2), we get:
$\frac{1}{3}\left(-\frac{4}{3} + \frac{10}{9}y\right) + \frac{y}{2} - \frac{13}{6} = 0$
$\implies -\frac{4}{9} + \frac{10}{27}y + \frac{y}{2} - \frac{13}{6} = 0$
$\implies \frac{10}{27}y + \frac{y}{2} = \frac{13}{6} + \frac{4}{9}$
$\implies \frac{20y + 27y}{54} = \frac{39 + 8}{18}$
$\implies 47y = 47 \times \frac{54}{18}$
$\implies y = 3$
Putting the value of y in (3), we get:
$\implies x = -\frac{4}{9} + \frac{10}{9} \times 3$
$\implies x = -\frac{4}{3} + \frac{10}{3} \implies x = \frac{6}{3} = 2$
$\implies x = 2, \quad y = 3$
Example: 2. Solve $2x + 3y = 11$ and $2x - 4y = -24$ and hence find the value of ‘m’ for which $y = mx + 3$ .
Solution: $2x + 3y = 11 \quad \dots \text{(1)}$
$2x - 4y = -24 \implies 2x = -24 + 4y$
$\implies x = \frac{-24 + 4y}{2}$
$\implies x = \frac{2(-12 + 2y)}{2}$
$\implies x = -12 + 2y \quad \dots \text{(2)}$
Putting the value of x in (1), we get:
$2(-12 + 2y) + 3y = 11$
$\implies -24 + 4y + 3y = 11$
$\implies 7y = 11 + 24$
$\implies y = \frac{35}{7} = 5$
Putting the value of y in (2), we get:
$x = -12 + 2 \times 5 = -12 + 10 = -2$
$\implies x = -2, \quad y = 5$

Class 10 Substitution Method

Example: 3. From the pair of linear equations for the following problems and find their solution by substitution method.
Example: 3 (i). The difference between two numbers is 26 and one number is three times the other. Find them.
Solution: Let the numbers are x and y.
$x - y = 26 \quad \dots \text{(1)}$
$x = 3y \quad \dots \text{(2)}$
Putting value of x from (2) to equation (1), we get:
$3y - y = 26$
$\implies 2y = 26$
$\implies y = \frac{26}{2} = 13$
Putting the value of y in (2), we get:
$\implies x = 3 \times 13 = 39$
$x = 39, \quad y = 13$
Example: 3 (ii). The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution: Let the angles are x, y.
$x - y = 18 \quad (x > y) \quad \dots \text{(1)}$
$x + y = 180 \quad \dots \text{(2)}$
From (1) $x = 18 + y \quad \dots \text{(3)}$
Putting the value of x in (2), we get:
$18 + y + y = 180 \implies 2y = 180 - 18$
$\implies y = \frac{162}{2} = 81$
Putting the value of y in (1), we get:
$x - 81 = 18 \implies x = 81 + 18$
$\implies x = 99, \quad y = 81$
Example: 3 (iii). The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
Solution: Let the cost of each bat and ball be x and y respectively.
$7x + 6y = 3800 \quad \dots \text{(1)}$
$3x + 5y = 1750 \quad \dots \text{(2)}$
From (2): $3x = 1750 - 5y \implies x = \frac{1750 - 5y}{3}$
Putting the value of x in (1), we get:
$7\left(\frac{1750 - 5y}{3}\right) + 6y = 3800$
$\implies \frac{12250 - 35y + 18y}{3} = 3800$
$\implies 12250 - 17y = 3800 \times 3$
$\implies -17y = 11400 - 12250$
$\implies y = \frac{-850}{-17} = 50$
$\implies y = 50$
Putting the value of y in $x = \frac{1750 - 5y}{3}$ , we get:
$x = \frac{1750 - 5 \times 50}{3} = \frac{1750 - 250}{3}$
$= \frac{1500}{3} = 500$
$\implies x = 500, \quad y = 50$
Example: 3 (iv). The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and charge per km? much does a person have to pay for travelling a distance of 25 km?
Solution: Let the fixed charge = x and charge per km = y
$x + 10y = 105 \quad \dots \text{(1)}$
$x + 15y = 155 \quad \dots \text{(2)}$
$x = 155 - 15y \quad \dots \text{(3)}$
Putting the value of x in (1), we get:
$(155 - 15y) + 10y = 105$
$\implies -5y = 105 - 155$
$\implies -5y = -50$
$\implies y = \frac{-50}{-5} = 10$
Putting the value of y in (3), we get:
$x = 155 - 15 \times 10 = 155 - 150 = 5$
$\implies x = 5, \quad y = 10$
Person have to pay for travelling a distance of 25 km = $25 \times 10 + 5 = 255$
Example: 3 (v). A fraction becomes $\frac{9}{11}$ , if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$ . Find the fraction.
Solution: Let the numerator of fraction = x, and denominator = y
Fraction= $\frac{x}{y}$
$\frac{x + 2}{y + 2} = \frac{9}{11}$
$\implies 11(x + 2) = 9(y + 2)$
$\implies 11x + 22 = 9y + 18$
$\implies 11x - 9y = 18 - 22$
$\implies 11x - 9y = -4 \quad \dots \text{(1)}$
Again:
$\frac{x + 3}{y + 3} = \frac{5}{6} \implies 6(x + 3) = 5(y + 3)$
$\implies 6x + 18 = 5y + 15$
$\implies 6x - 5y = 15 - 18$
$\implies 6x - 5y = -3 \quad \dots \text{(2)}$
$\implies 6x = -3 + 5y$
$\implies x = \frac{-3 + 5y}{6} \quad \dots \text{(3)}$
Putting the value of x in (1), we get:
$\implies 11\left(\frac{-3 + 5y}{6}\right) - 9y = -4$
$\implies \frac{-33 + 55y - 54y}{6} = -4$
$\implies -33 + y = -24$
$\implies y = 33 - 24 = 9$
Putting the value of y in (3), we get:
$x = \frac{-3 + 5 \times 9}{6} = \frac{-3 + 45}{6} = \frac{42}{6} = 7$
$\implies x = 7, \quad y = 9$
fraction= $\frac{7}{9}$
Example: 3 (vi). Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution: Let present age of Jacob = x, present age of Jacob’s son = y.
Five years ago Jacob’s Age = x – 5
Five years ago Age of his son = y – 5
$x - 5 = 7(y - 5) \implies x - 5 = 7y - 35$
$\implies x - 7y = -35 + 5$
$x - 7y = -30 \quad \dots \text{(1)}$
After five years:
Jacob’s Age = x + 5
Age of his son = y + 5
$x + 5 = 3(y + 5)$
$\implies x + 5 = 3y + 15$
$\implies x = 3y + 15 - 5$
$x = 3y + 10 \quad \dots \text{(2)}$
Putting the value of x in (1), we get:
$3y + 10 - 7y = -30$
$\implies -4y = -30 - 10 \implies y = \frac{-40}{-4} = 10$
Putting the value of y in (2), we get:
$x = 3 \times 10 + 10 = 30 + 10 = 40$
Age of Jacob = 40, Age of his son = 10
By solving the above examples, one can understand Class 10 Substitution Method.

3.Practice Problems of Class 10 Substitution Method for Students:

Class 10 Substitution Method

Solve the following pair of linear equations by the substitution
(1.)5x-2y=19,3x+y=18
Answers:(1)x=5,y=3
By solving the above questions,you can understand the Class 10 Substitution Method well because the concept is well understood when you solve it practically.
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Also Read This Article:- Qudratic Polynomials in Class 10: Relationship Between Zeroes and Coefficients

4.Frequently Asked Questions Related to Class 10 Substitution Method:

Q:1.What is the substitution method?

Ans:We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations.That is the method known as the substitution method.

Q:2.Write the working rule of substitution method

Ans:Step:1.Find the one variable,say y in terms of the other variable i.e.,x from either equation,whichever is convenient.
Step:2.Substitute this value of y in the other equation,and reduce it to an equation in one variable,i.e., in terms of x,which can be solved.If this statement is true,you conclude that the pair of linear equations has infinitely many solutions.If the statement is false,then the pair of linear equations is inconsistent
Step:3.Substitute the value of x (or y) obtained in step 2 in the equation used in step 1 to obtain the value of the other variable.

Q:3.What is meant by algebraic methods?

Ans:Algebraic methods
(1.)Substitution Method
(2.)Elimination Method
(3.)Cross-multiplication Method
By answering the above questions,you can know about the primary terms of Class 10 Substitution Method.

5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

Class 10 Substitution Method

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*”कौनसी संख्या दोनों प्रश्नचिह्नों के स्थान पर आएगी?”*
(Which number will come in place of both the question marks?)
$\frac{?}{32} = \frac{72}{?}$
*पिछली प्रश्नोत्तरी का हल*
$5\frac{3}{4} + 2\frac{6}{8} = 11\frac{8}{9} - ?$
$? = (11 - 7) + \left(\frac{8}{9} - \frac{3}{4} - \frac{6}{8}\right)$
$= 4 + \left(-\frac{11}{18}\right) = 3\frac{7}{18}$
*Previous Quiz Solution*
$5\frac{3}{4} + 2\frac{6}{8} = 11\frac{8}{9} - ?$
$\implies ? = -5\frac{3}{4} + 2\frac{3}{4} + 11\frac{8}{9}$
$= (11 - 5 - 2) + \left(\frac{8}{9} - \frac{3}{4} - \frac{3}{4}\right)$
$= 4 + \left(\frac{32 - 27 - 27}{36}\right)$
$= 4 + \left(-\frac{22}{36}\right)$
$= 4 - \frac{11}{18}$
$= \frac{72 - 11}{18} = \frac{61}{18}$
$= 3\frac{7}{18}$
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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Sanjay Kumawat

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