The cross-multiplication method is a powerful and straightforward technique used to solve a pair of linear equations. A pair of linear equations can be solved through a variety of methods. However the “Cross-Multiplication method” stands out for its speed and efficiency. Not only it is used in solving a pair of linear equations but also it has numerous applications e.g., solving matrix algebra, vector calculus, analytical geometry, statistics etc.

In this article, we will discuss the cross-multiplication method and solve several linear equations using this method. So let’s get started!

What is Cross-Multiplication Method?

Cross-multiplication method is the fastest technique to solve systems of linear equations in two variables. This method is only applicable when a pair of linear equations in two variables is given to us. Learn more about Cross-Multiplication Method :

Formula of Cross-Multiplication Method

Let there be an pair of linear equations given by:

a₁x + b₁y + c_{1 = 0}

a₂x + b₂y + c_{2 = 0}

..

Cross Multiplication Method : Practice Questions with Solution

Let’s discuss few questions on the basis of the formula of cross multiplication method.

Question 1. Solve the pair of linear equations given below :

2x – y = -1

3x + 2y = 9

Solution:

Given equations are:

• 2x – y = -1
• 3x + 2y = 9

Arranging these equations in the standard form ax + by + c = 0 , we get –

2x – y + 1 = 0 and 3x + 2y – 9 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 2, b₁ = -1, c₁ = 1

a₂ = 3, b₂ = 2, c2 = -9

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

x/[(-1)(-9) – (2)(1)] = y/[(1)(3) – (-9)(2)] = 1/[(2)(2) – (3)(-1)]

x/7 = y/21 = 1/7

x/7 = 1/7 and y/21 = 1/7

x = 1 and y = 3

Question 2: Solve for x and y:

3x + 2y = 11

6x – 4y = 14

Solution:

Given equations are:

3x + 2y = 11

6x + 4y = 14

Arranging these equations in the standard form ax + by + c = 0 , we get –

3x + 2y – 11 = 0 and 6x + 4y – 14 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 3, b₁ = 2, c₁ = -11

a₂ = 6, b₂ = 4, c₂ = -14

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Here we see that the denominator becomes zero on putting the values of the coefficients. So this pair of equations has no solution.

Question 3: Solve the system of equations:

2x + 5y = 13

3x – 2y = 4

Solution:

Given equations are:

2x + 5y = 13

3x – 2y = 4

Arranging these equations in the standard form ax + by + c = 0 , we get –

2x + 5y – 13 = 0 and 3x – 2y – 4 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

a₁ = 2, b₁ = 5, c₁ = -13

a₂ = 3, b₂ = -2, c₂ = -4

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

x/[(5)(-4) – (-2)(-13)] = y/[(-13)(3) – (-4)(2)] = 1/[(2)(-2) – (3)(5)]

x/(-46) = y/(-31) = 1/(-19)

x/(-46) = 1/(-19) and y/(-31) = 1/(-19) => x = 46/19 and y = 31/19

Question 4: Solve the system of linear equations:

2x + 3y = 7

x – 3y = -3

Solution:

Given equations are:

2x + 3y = 7

x – 2y = -3

Arranging these equations in the standard form ax + by + c = 0 , we get –

2x + 3y – 7 = 0 and x – 2y + 3 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 2, b₁ = 3, c₁ = -7

a₂ = 1, b₂ = -2, c₂ = 3

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

x/[(3)(3) – (-2)(-7)] = y/[(-7)(1) – (3)(2)] = 1/[(2)(-2) – (1)(3)]

x/(-5) = y/(-13) = 1/(-7)

x/(-5) = 1/(-7) and y/(-13) = 1/(-7) => x = 5/7 and y = 13/7

Question 5: Find the solution to the pair of equations:

x – 3y = 2

2x + 2y = 8

Solution:

Given equations are:

x – 3y = 2

2x + 2y = 8

Arranging these equations in the standard form ax + by + c = 0 , we get –

x – 3y -2 = 0 and 2x + 2y – 8 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 1, b₁ = -3, c₁ = -2

a₂ = 2, b₂ = 2, c₂ = -8

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

x/[(-3)(-8) – (2)(-2)] = y/[(-2)(2) – (1)(-8)] = 1/[(1)(2) – (2)(-3)]

x/(28) = y/(4) = 1/(8)

x/(28) = 1/(8) and y/(4) = 1/(8) => x = 7/2 and y = 1/2

Question 6: Solve the system of equations:

x + y = 14

x – y = 4

Solution:

Given equations are:

x + y = 14

x – y = 4

Arranging these equations in the standard form ax + by + c = 0 , we get –

x + y – 14 = 0 and x – y – 4 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 1, b₁ = 1, c₁ = -14

a₂ = 1, b₂ = 1, c₂ = -4

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

x/[(1)(-4) – (-14)(-1)] = y/[(-14)(1) – (-4)(1)] = 1/[(1)(-1) – (1)(1)]

x/(-18) = y/(-10) = 1/(-2)

x/(-18) = 1/(-2) and y/(-10) = 1/(-2) => x = 9 and y = 5

Question 7: Solve for x and y

3x/2 – 5y/3 = -2

x/3 + y/2 = 13/6

Solution:

Given equations are:

3x/2 – 5y/3 = -2

x/3 + y/2 = 13/6

Arranging these equations in the standard form ax + by + c = 0 , we get –

9x – 10y + 12 = 0 and 2x + 3y – 13 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 9, b₁ = -10, c₁ = 12

a₂ = 2, b₂ = 3, c₂ = -13

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

x/[(-10)(-13) – (-12)(3)] = y/[(2)(12) – (9)(-13)] = 1/[(9)(3) – (2)(-10)]

x/(94) = y/(-141) = 1/(47)

x/(94) = 1/(47) and y/(-141) = 1/(47) => x = 2 and y = 3

Question 8: Find the solution to the pair of equations:

s – t = 3

s/3 + t/2 = 6

Solution:

Given equations are:

s – t = 3

s/3 + t/2 = 6

Arranging these equations in the standard form as + bt + c = 0 , we get –

s – t – 3 = 0 and 2s + 3t – 36 = 0

Comparing with a₁s + b₁t + c₁ = 0 and a₂s + b₂t + c₂ = 0,

a₁ = 1, b₁ = -1, c₁ = -3

a₂ = 2, b₂ = 3, c₂ = -36

Using the cross multiplication formula,

s/(b₁c₂ – b₂c₁) = t/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

s/[(-1)(-36) – (3)(-3)] = t/[(-3)(2) – (1)(-36)] = 1/[(1)(3) – (2)(-1)]

s/(45) = t/(30) = 1/(5)

s/(45) = 1/(5) and t/(30) = 1/(5) => s = 9 and t = 6

Question 9: Solve the pair of linear equations given below:

9x – 4y = 2

7x – 3y = 2

Solution:

Given equations are:

9x – 4y = 2

7x – 3y = 2

Arranging these equations in the standard form ax + by + c = 0 , we get –

9x – 4y – 2 = 0 and 7x – 3y – 2 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 9, b₁ = -4, c₁ = -2

a₂ = 7, b₂ = -3, c₂ = -2

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

x/[(-4)(-2) – (-3)(-2)] = y/[(-2)(7) – (9)(-2)] = 1/[(9)(-3) – (7)(-4)]

x/2 = y/4 = 1/1

x/2 =1/1 and y/4 = 1/1

x = 2 and y = 4

Question 10: Solve the pair of linear equations:

x + y = 5

2x – 3y = 4

Solution:

Given equations are:

x + y = 5

2x – 3y = 4

Arranging these equations in the standard form ax + by + c = 0 , we get –

x + y – 5 = 0 and 2x – 3y – 4 = 0

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 1, b₁ = 1, c₁ = -5

a₂ = 2, b₂ = -3, c₂ = -4

Using the cross multiplication formula,

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

Putting the values, we get-

x/[(1)(-4) – (-3)(-5)] = y/[(-5)(2) – (1)(-4)] = 1/[(1)(-3) – (2)(1)]

x/(-19) = y/(-6) = 1/(-5)

x/(-19) = 1/(-5) and y/(-6) = 1/(-5)

x = 19/5 and y = 6/5

Cross-Multiplication Method: Worksheet

Worksheet on cross multiplication method is added in form of image below:

Answer Key:

1. x = 3, y = 2
2. x = 2, y = 3
3. x = 3, y = 2
4. x = 3, y = 2
5. x = 3, y = 2
6. x = 3, y = 2
7. x = 4, y = 2
8. x = 3, y = 2
9. x = 4, y = 2
10. x = 5, y = 3

Related Articles:

• Cross Multiplication Method: Formula, Derivation, Examples
• Pair Of Linear Equations In Two Variables
• Sequence and Series : Definition, Types, Formulas

Frequently Asked Questions

What is Cross-Multiplication Method?

It is a technique to solve equations of the form ax + by + c = 0, where a, b and c are constants.

Can I use cross-multiplication method for systems of linear equations:

Yes, You can apply the method to each pair of equation.

Is there any other method to solve a pair of linear equation?

There are several method to solve a pair of linear equation:

• Substitution Method
• Elimination Method
• Graphical Method
• Matrix Method
• Cramer’s Rule
• Augumented Matrix Method