Substitution Method Questions

Substitution Method questions are presented here to solve the system of equations faster. We know that the substitution method is a kind of algebraic method to solve the simultaneous linear equations involving two variables. The substitution method questions provided here are framed as per the CBSE and NCERT Curriculum. Each question is provided with a complete explanation, so that you can cross verify your answers with the solutions on our page. To learn more about substitution methods, click here.

Substitution Method Questions with Solutions

1. Solve the system of equations using substitution method:

a + b = 0

a – b = 2.

Also, verify your answer.

Solution:

Given system of equations:

a + b = 0 …(1)

a – b = 2 …(2)

Step 1: Finding the value of “a” in terms of “b”.

Hence, from equation (1), we get

a = -b …(3)

Step 2: Now, substitute a = -b in equation (2)

⇒ (-b) – b = 2

⇒ -2b = 2

⇒ b = -1

Step 3: substituting b = -1 in (1), we get

⇒ a -1 = 0

⇒ a = 1

Hence, the values of a and b are 1 and -1 respectively.

Verification:

To check whether the values of a and b are correct, substitute the values in any of the given equations.

So, put a = 1 and b = -1 in equation (2),

1 – (-1) = 2

1 + 1 = 2

2 = 2

Hence, LHS = RHS.

2. Solve the equations 2p – 3q = 2 and p + 2q = 8, using the method of substitution.

Solution:

Given:

2p – 3q = 2 …(1)

p + 2q = 8 …(2)

Using the substitution method, we can find the values of p and q as given below:

From equation (1),

⇒ 2p = 2 + 3q

⇒ p = (2 + 3q)/2

Now, substituting p = (2 + 3q)/2 in (2), we get

⇒ [(2 + 3q)/2] + 2q = 8

⇒ 2 + 3q + 4q = 16

⇒ 2 + 7q = 16

⇒7q = 16 – 2

⇒ 7q = 14

⇒ q = 14/7 = 2

Putting q = 2 in equation (2),

p + 2(2) = 8

p + 4 = 8

p = 8 – 4

p = 4

Hence, the solution to the system of equations 2p – 3q = 2 and p + 2q = 8 are:

p = 4 and q = 2.

3. Solve the system of linear equations using substitution method:

p + q = 14 and p – q = 2

Solution:

Given:

p + q = 14 …(1)

p – q = 2 … (2)

From equation (1),

p = 14 – q …(3)

Now, substitute (3) in (2), we get

(14 – q) – q = 2

14 – 2q = 2

-2q = 2 – 14

-2q = -12

q = -12/-2

q = 6

Now, substitute q = 6 in (1)

p + 6 = 14

p = 14 – 6

p = 8

Hence, the solutions are: p = 8 and q = 6.

4. Solve the pair of linear equations given below using the substitution method:

3p – q = 3

9p – 3q = 9

Solution:

Given linear equations:

3p – q = 3 …(1)

9p – 3q = 9 …(2)

From (1), we can write

⇒ 3p = 3 + q

⇒ p = (3+q)/3 …(3)

Now, substitute (3) in (2), we get

⇒ 9[(3+q)/3] – 3q = 9

⇒ 9 +3q – 3q = 9

⇒ 9 = 9

Therefore, substituting different values of q in p = (3+q)/3, we get different p values.

Thus, the system of linear equations has infinitely many solutions.

5. Create a pair of linear equations from the given statement, and determine the solution using the substitution method.

“Find the numbers if one number is three times of the other number. Also, given that the difference between two numbers is 26”.

Solution:

From the given condition, the equations formed are:

p = 3q …(1)

p – q = 26 ..(2) [Assuming that p and q are the two numbers]

Here, equation (1) is already represented in terms of the variable “q”.

So, substituting equation (1) in (2), we get

(3q) – q = 26

2q = 26

q = 26/2

q = 13

Now, plug the value of q in equation (1), to get the value of p.

⇒ p = 3(13)

⇒ p = 39

Therefore, the two numbers are 13 and 39.

Also, check:

• Elimination Method
• Cross Multiplication Method
• Graphing of Linear Equations
• Linear Equations in Two Variables

6. Using substitution method, solve the following equations:

p + 3q = 13 and 3p + q = 7.

Also, justify your answer.

Solution:

Given equations:

p + 3q = 13 …(1)

3p + q = 7 …(2)

From equation (1), we can write

p = 13 – 3q ..(3)

Substituting (3) in (2), we get

⇒ 3(13 – 3q) + q = 7

⇒ 39 – 9q + q = 7

⇒ 39 – 8q = 7

⇒ -8q = 7 – 39

⇒ -8q = -32

⇒ 8q = 32

⇒ q = 32/8

⇒ q = 4

Now, substituting q = 4 in equation (1), we get

⇒ p + 3(4) = 13

⇒ p + 12 = 13

⇒ p = 13 – 12

⇒ p = 1

Hence, the solutions are:

p = 1 and q = 4

Verification:

Substitute p = 1 and q = 4 in equation (1),

⇒ 1 + 3(4) = 13

⇒ 1 + 12 = 13

⇒ 13 = 13

Thus, LHS = RHS.

Therefore, p = 1 and q = 4 are the solutions of the equations p + 3q = 13 and 3p + q = 7.

7. Solve the linear equations using the substitution method:

11p + 15q = -23

7p – 2q = 20

Solution:

Given system of linear equations are:

11p + 15q = -23 …(1)

7p – 2q = 20 …(2)

From equation (1), we can write the following:

⇒ 11p = -23 – 15q

⇒ 11p = – (23 + 15q)

⇒ p = – (23 + 15q)/11 …(3)

Now, substitute equation (3) in (2)

⇒ 7[- (23 + 15q)/11] – 2q = 20

⇒ -(161 + 105q) – 22q = 220

⇒ -161 – 105q – 22q = 220

⇒ -105q – 22q = 220 + 161

⇒ – 127q = 381

⇒ q = -381/127

⇒ q = -3

Substituting q = -3 in (2),

⇒ 7p – 2(-3) = 20

⇒ 7p + 6 = 20

⇒ 7p = 20 – 6

⇒ 7p = 14

⇒ p = 14/7

⇒ p = 2

Therefore, the values of p and q are 2 and -3, respectively.

8. Find the values of x and y from the system of linear equations:

7p + 6q = 3800, and 3p + 5q = 1750

Solution: p = 500, q = 50

Given linear equations:

7p + 6q = 3800 .. (1)

3p + 5q = 1750 …(2)

From (1), we can write:

7p = 3800 – 6q

⇒ p = (3800 – 6q)/7 …(3)

Substituting (3) in (2),

⇒ 3[(3800 – 6q)/7] + 5q = 1750

⇒ 11400 – 18q + 35q = 12250

⇒ -18q + 35q = 12250 – 11400

⇒ 17q = 850

⇒ q = 850/17

⇒ q = 50

Now, plug the value of q in equation (1), we get

7p + 6(50) = 3800

7p + 300 = 3800

7p = 3800 – 300

7p = 3500

p = 3500/7

p = 500

Hence, the values of p and q are 500 and 50, respectively.

9. Find the values of p and q using the method of substitution:

p + 3q = 10

2p + 3q = 21

Also, justify your answer.

Solution:

Given equations:

p + 3q = 10 …(1)

2p + 3q = 21 …(2)

Now, equation (1) can be written as follows:

p = 10 – 3q …(3)

Substituting equation (3) in (2), we get

2(10 – 3q) + 3q = 21

20 – 6q + 3q = 21

20 – 3q = 21

-3q = 21 – 20

-3q = 1

Hence, q = -⅓.

Now, plug the value q = -⅓ in equation (1), we get

p + 3(-⅓) = 10

p – 1 = 10

p = 10 + 1

p = 11.

Hence, the values of p and q are 11 and -⅓, respectively.

Verification:

Substitute p = 11 and q = -⅓ in equation (1)

11 + 3(-⅓) = 10

11 – 1 = 10

10 = 10

Hence, LHS = RHS.

10. Solve the equations using substitution method and find the values of a and b:

0.2a + 0.3b = 1.3

0.4a + 0.5b = 2.3.

Also, verify your answer.

Solution:

Given:

0.2a + 0.3b = 1.3 …(1)

0.4a + 0.5b = 2.3 … (2)

From equation (1), we get

0.2a = 1.3 – 0.3b

Hence, a = (1.3 – 0.3b)/0.2 …(3)

Substituting equation (3) in (2), we get

0.4[(1.3 – 0.3b)/0.2] + 0.5b = 2.3

0.4(1.3 – 0.3b) + 0.1b = 0.46

0.52 – 0.12b + 0.1b = 0.46

– 0.12b + 0.1b = 0.46 – 0.52

-0.02b = -0.06

Cancelling minus sign on both sides, we get

0.02b = 0.06

Hence, b = 3.

Now, substitute b = 3 in equation (1), we get

0.4a + 0.5(3) = 2.3

0.4a + 1.5 = 2.3

0.4a = 2.3 – 1.5

0.4a = 0.8

a = 0.8 /0.4

a = 2

Hence, a = 2 and b = 3 are the solutions of the equations 0.2a + 0.3b = 1.3 and 0.4a + 0.5b = 2.3.

Verification:

Substitute a = 2 and b = 3 in equation (1).

⇒ 0.2(2) + 0.3(3) = 1.3

⇒ 0.4 + 0.9 = 1.3

⇒ 1.3 = 1.3

Thus, LHS = RHS.

Hence, a = 2 and b = 3 are the solutions to the linear equation 0.2a + 0.3b = 1.3 and 0.4a + 0.5b = 2.3.

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Practice Questions

Answer the following questions:

1. Compute the values of and b using substitution method: √2 a + √3 b = 0 and √3 a – √8 b = 0

2. Solve the equations 2p + 3q = 11 and 2p – 4q = – 24 using the substitution method. Also, determine the value of “m”, such that y = mx + 3.

3. Find the values of “r” and “s”: r + 10s = 105 and r + 15s = 155.

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