Simultaneous Equations Questions

Simultaneous equations questions with solutions are given here for practice and understanding how to determine the solution of given equations. Simultaneous equations are the system of two or more independent equations which satisfies a common solution. Simultaneous equations could have,

• Linear Equation
• Quadratic Equation
• Polynomial Equations

There are several methods of solving simultaneous equations, and they are:

If we are given a pair of linear equations, the existence of the solution is determined as follows:

Learn more about Simultaneous Equations.

Video Lesson on Consistent and Inconsistent Simultaneous Equations

Simultaneous Equations Questions with Solutions

Let us solve a few simultaneous equations questions.

Question 1:

Solve the following:

2x + 3y = –2

5x + 4y + 2 = 0

Solution:

Given,

2x + 3y = –2 ….(i)

5x + 4y = –2 ….(ii)

Multiply (i) by 5 and (ii) 2 on both sides, subtracting (ii) from (i), we get

10x + 15y = –10

10x + 8y = –4

(–)_ (–)___(+)___

7y = –6

⇒ y = –6/7

Substituting the value of y in (i), we get

x = 2/7

∴ the solution of the given simultaneous equation is (2/7, –6/7)

Question 2:

Solve the following equations:

2a – b = 4 and 2b – 3a = 2

Solution:

Given,

2a – b = 4 …..(i)

2b – 3a = 2

⇒ –3a + 2b = 2 ….(ii)

From (i),

a = ½ (b + 4)

Putting this value in (ii)

–3/2(b + 4) + 2b = 2

⇒ –3b – 12 + 4b = 4

⇒ b = 16

Thus, a = ½(16 + 4) = 10

∴ the solution of the given simultaneous equation is a = 10 and b = 16.

Question 3:

Solve the following simultaneous equations:

7m – 9n = 3 and 2m – 3n = 1

Solution:

Given,

7m – 9n = 3 ….(i)

2m – 3n = 1 ….(ii)

Multiplying equation (ii) by 3 on both sides, then subtracting from (i), we get,

(7m – 6m) + ( –9n + 9n) = 3 – 3

⇒ m = 0

Consequently, n = –⅓

∴ the solution of the given simultaneous equation is m = 0 and n = –⅓.

Question 4:

Solve the following simultaneous equations:

a² + 2b = 9 and b – a = 3

Solution:

Given,

a² + 2b = 9 …(i)

b – a = 3 …(ii)

From (ii) b = 3 + a

Substituting the value of b in (i), we get

a² + 2(3 + a) = 9

⇒ a² + 2a + 6 = 9

⇒ a² + 2a + – 3 = 0

⇒ (a + 3)(a – 1) = 0

⇒ a = 1 and –3

When a = 1, b = 4, and when a = –3, b = 0

∴ the solution of the given simultaneous equation is a = 1, b = 4 and a = –3, b = 0.

Also, learn about solving simultaneous equations by Cramer’s rule.

Question 5:

Solve the following simultaneous equations:

a² – b = 14 and 2b – 4 = 12a

Solution:

Given,

a² – b = 14 ….(i)

2b – 4 = 12a

⇒ 12a – 2b = –4

⇒ 6a – b = –2

⇒ b = 6a + 2

Substituting the value of b in (i), we get

a² – 6a – 2 = 14

⇒ a² – 6a – 16 = 0

⇒ (a – 8)(a + 2) = 0

⇒ a = 8 and –2

When a = 8, b = 50 and when a = –2, b = 10

∴ the solution of the given simultaneous equation is a = 8, b = 50 and a = –5, b = 11.

Question 6:

Solve the following simultaneous equations:

8q + p + r = 0

q + 2p + r = 0

q + p – 1600 = 0

Solution:

Given,

8q + p + r = 0 …(i)

q + 2p + r = 0 …(ii)

q + p – 1600 = 0 …(iii)

Subtracting (ii) from (i), we get

7q – p = 0 ….(iv)

Solving (iii) and (iv), we get,

q = 200 and p = 1400

And from (i), we get r = –3000

Question 7:

Solve the following equations:

(m + n – 8)/2 = (m + 2n – 14)/3 = (3m + n – 12)/11

Solution:

Taking the first two equations:

(m + n – 8)/2 = (m + 2n – 14)/3

⇒ 3m + 3n – 24 = 2m + 4n – 28

⇒ m – n = – 28 + 24

⇒ m – n = –4 ….(i)

Taking the last two equations:

(m + 2n – 14)/3 = (3m + n – 12)/11

⇒ 11m + 22n – 154 = 9m + 3n – 36

⇒ 2m + 19n = 154 – 36

⇒ 2m + 19n = 118 ….(ii)

Multiplying both sides of (i) by 2 and then subtracting from (ii), we get

19n + 2n = 118 + 8

⇒ 21n = 126

⇒ n = 126/21 = 6

And m = 2

∴ the solution of the given simultaneous equation is m = 2 and n = 6.

The other methods of solving simultaneous equations are:

• Gaussian Elimination
• Jacobi and Gauss-Seidel Iterative Methods

Question 8:

A man can row 16 km downstream and 8 km upstream in 6 hours. He can row 6 km upstream and 24 km downstream in 6 hours. Find the speed of man in still water.

Solution:

Let the speed of a man rowing a boat in still water be x km/hr and the speed of the stream be y km/hr.

Speed of rowing in downstream = x + y = 1/u (let)

Speed of rowing in upstream = x – y = 1/v

Now, given

16/(x + y) + 8/(x – y) = 6

⇒ 16u + 8v = 6

⇒ 8u + 4v = 3 ….(i)

And 24/(x + y) + 6/(x – y) = 6

⇒ 24u + 6v = 6

⇒ 4u + v = 1 ….(ii)

Solving (i) and (ii) we get, u = ⅛ ⇒ x + y = 8 and v = ½ ⇒ x – y = 2

Thus, we get x = 5km/hr and y = 3km/hr.

∴ The speed of the man rowing the boat in still water is 5 km/hr.

Question 9:

If (sin 𝜃, cos 𝜃) satisfies the system of equations mx + ny + a + b = a – b and nx + my + 2b = 0, then find the value of 𝜃 where 0≤ 𝜃 ≤ 90^o.

Solution:

Given equations,

mx + ny + a + b = a – b

⇒ mx + ny + 2b = 0 …..(i)

And nx + my + 2b = 0 ….(ii)

Multiply (i) by n and (ii) by m on both sides, subtracting (ii) from (i), we get,

(n² – m²)y + 2nb – 2mb = 0

y = –2/(m + n)

Similarly, x = –2/(m + n)

That is, x = y ⇒ sin 𝜃 = cos 𝜃, for 𝜃 = 45^o, the value of sin 𝜃 is equal to the value of cos 𝜃.

Question 10:

A sum of ₹ 400 was distributed among the students of a class. Each boy received ₹ 8, and each girl received ₹ 4. If each girl had received ₹ 10, then each boy would have received ₹ 5. Find the total number of students in the class.

Solution:

Let the number of boys be x, and the number of girls be y. Then

8x + 4y = 400 and 5x + 10y = 400

⇒ 2x + y = 100 and x + 2y = 80

Multiplying the second equation by 2 on both sides, then subtracting from the first equation, we get,

–3y = –60 ⇒ y = 20 and x = 40

∴ there are 40 boys and 20 girls in the classroom.

Practice Questions on Simultaneous Equations

1. Solve the following simultaneous equations:

(i) 3x + 5y = 6 and 2x + 6y = –2

(ii) 2x + y = z, x + 4y = 3z and x – 3y = 16

(iii) 2x² – 1 = 5 and x + y = –2

(iv) x/a + y/b = a² + b² and x/a² + y/b² = a + b

2. Four years ago, the age of a person was thrice that of his son. Eight years later, the age of the person will be twice that of his son. Find the present ages of the person and his son.

3. For what value of k, the following simultaneous equations will not have any solutions: 2x + 3y = 1 and (3k – 1)x + (1 – 2k)y = 2k + 3?

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