Cross Multiplication Method: Formula, Derivation & Examples

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Cross multiplication method is used when solving linear equations with two variables. The method of cross-multiplication is the simplest and most straightforward method of solving linear equations in two variables.

This approach is commonly used when there are two variables in a linear equation.
The solution of linear equations is divided into graphical and algebraic methods.
Algebraic method is divided into elimination method, substitution method and cross-multiplication method.
In the cross multiplication method, the numerator on the left-hand side is multiplied by the denominator on the right-hand side.
The method is also used to determine whether two linear equations are equivalent.
It can be used to compare fractional values.
The condition that needs to be satisfied in the case of the cross-multiplication method is given by:

b₂a₁ – b₁a₂ ≠ 0

Table of Content

Cross Multiplication Method
Cross Multiplication Formula
Derivation of Cross Multiplication Method
Unique Solution by Cross Multiplication Method
Things to Remember
Sample Questions

Key Terms: Cross Multiplication Method, Multiplication, Linear Equations, Coefficients, Variables, Substitution Method, Elimination Method, Numeraotor, Denominator, Unique Solutions, Formula Cross Multiplication Method

Cross Multiplication Method

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Cross-multiplication is a method for determining the solution of two-variable linear equations. It is the quickest approach for solving a pair of linear equations.

For a given pair of two-variable linear equations:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

The values x and y will be as follows when cross multiplication is used:

Cross Multiplication Method

Example of Cross Multiplication Method

Example. Using the cross-multiplication method, find the value of x and y: 4x – 3y – 6 = 0, 3x + 4y – 17 = 0

Ans: Two equations are given,

3x + 4y -17 = 0
4x - 3y - 6 = 0

By the method of cross-multiplication,

\(\frac{x}{4(-6) - {(-3)(-17)}}\)=\(\frac{y}{(-17)4 - (-6)3}\)=\(\frac{1}{3(-3) - (4*4)}\)

\(\frac{x}{(-75)}\)=\(\frac{y}{-50}\)=\(\frac{1}{-25}\)

Multiply by 25

\(\frac{x}{3}\)=\(\frac{y}{2}\)=1

Hence, required solution: x = 3, y = 2.

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Cross Multiplication Formula

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The cross-multiplication formula is used to solve linear equations with two variables:

Cross Multiplication Method Formula

Example of Cross Multiplication Formula

Example. Using the cross-multiplication method, find the value of x and y: 3x – 2y – 1 = 0, 2x + 3y – 7 = 0

Ans: Two equations are given,

3x - 2y -1 = 0
2x + 3y - 7 = 0

By the method of cross-multiplication,

\(\frac{x}{-2(-7) - {(3)(-1)}}\)=\(\frac{y}{(-7)3 - (-1)2}\)=\(\frac{1}{(3)(3) - (-2*2)}\)

\(\frac{x}{(-11)}\)=\(\frac{y}{-19}\)=\(\frac{1}{13}\)

Multiply by 13

Hence, required solution: x = -13/11, y = -13/19.

Derivation of Cross Multiplication Formula

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Consider any pair of linear equations to further comprehend this technique (that is, with any coefficients).

a₁x+b₁y+c₁=0

a₂x+b₂y+c₂=0

Let's write the coefficients in the original pair of equations in a grid format as follows.

a₁ b₁ c₁

a₂ b₂ c₂

We'll just disregard the x coefficients in our grid now.
Cross-multiply and subtract the coefficients in the remaining two columns:

Hence, the first part of the solution is given as

\(\frac{x}{b_1c_2 - b_2c_1}\)

Now, we have to consider the expression below -y, given as a₁c₂ - a₂c₁ ,
To write this, we disregard the y coefficients column and cross-multiply and subtract the coefficients in the remaining two columns:

2nd Picture

Second part of the solution is given as:

\(\frac{-y}{a_1c_2 - a_2c_1}\)

Finally, we evaluate the equation a₁b₂ - a₂b₁ (below 1).
To write this, we disregard the constants column and cross-multiply.
Subtract the coefficients in the remaining two columns:

3rd Picture

Therefore, the last portion of our equation becomes

\(\frac{1}{a_1b_2 - a_2b_1}\)

When we combine all three parts, we get the following complete solution to the pair of linear equations:

Cross Multiplication Method

Unique Solution by Cross Multiplication Method

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We can determine a unique result, inconsistently with infinitely many solutions while solving linear equations in two variables using the cross multiplication approach.

Let's go over the important points one by one.

a₁x+b₁y+c₁=0

a₂x+b₂y+c₂=0

Unique Solution

For obtaining a unique solution,

\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

Hence,

\(a_1b_1\neq a_2b_2\)

\(a_1b_1 - a_2b_2 \neq 0\)
This means that if we write the solution equivalence using the cross-multiplication technique.
There will be a unique solution as long as the term below 1 is non-zero:

4th pic

Inconsistent Solution

For obtaining a inconsistent solution, the required condition is given as:

a₁/a₂=b₁/b₂≠c₁/c₂

Inconsistent solution equation can be represented as:

a₁b₂ – a₂b₁ = 0

b₁c₂ – b₂c₁ ≠ 0

a₁c₂ – a₂c₁ ≠ 0

Infinitely Many Solution

For obtaining a infinitely many solution, the required condition is given as:

a₁/a₂=b₁/b₂=c₁/c₂

Infinitely many solution equations can be represented as:

a₁b₂ – a₂b₁ = 0

b₁c₂ – b₂c₁ = 0

a₁c₂ – a₂c₁ = 0

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Things to Remember

Cross multiplication method is used when solving linear equations with two variables.
The method of cross-multiplication is the simplest and most straightforward method of solving linear equations in two variables.
When there are two variables in a linear equation, this approach is commonly used.
The equation can be solved for unique, inconsistent and infinitely many solutions.
It is used to simplify the solution and check for equivalency of equations.

Sample Questions

Ques. Using the cross-multiplication method, solve the following linear equations. (3 marks)
3x-4y=2
y-2x=7

Ans. We are rewriting the above equations as

3x - 4y = 2

-2x + y =7

Using the method of cross-multiplication,

\(\frac{x}{b_1c_2 - b_2c_1}\)=\(\frac{b}{c_1a_2 - c_2a_1}\)=\(\frac{1}{a_1b_2 - a_2b_1}\)

Therefore, substituting values in the above equation

\(\frac{x}{28 + 2}\)=\(\frac{y}{4 + 21}\)=\(\frac{1}{3 - 8}\)

\(\frac{x}{30}\)=\(\frac{y}{25}\)=\(\frac{1}{5}\)

Hence,

x = -6 ; y = -5

Ques. Determine the variable values that satisfy the following equation:
2x + 5y = 20 and 3x+6y =12 (3 marks)

Ans. We know that,

2x + 5y = 20

3x + 6y = 12

By the cross-multiplication method,

\(\frac{x}{b_1c_2 - b_2c_1}\)=\(\frac{b}{c_1a_2 - c_2a_1}\)=\(\frac{1}{a_1b_2 - a_2b_1}\)

Therefore, substituting values in the above equation

\(\frac{x}{(5*12) - (6*20)}\)=\(\frac{y}{(20*3) - (12*2)}\)=\(\frac{1}{(5*3) - (6*2)}\)

\(\frac{x}{(-60)}\)=\(\frac{y}{36}\)=\(\frac{1}{3}\)

Here,

x = – 20

y = 12

As a result, the place where the given equations intersect is x = -20 and y = 12.

Ques. Using the cross-multiplication method, find the value of x and y (3 marks)
8x + 5y = 11
3x – 4y = 10

Ans. After transposition,

8x + 5y – 11 = 0

3x – 4y – 10 = 0

Therefore, substituting values in the above equation

\(\frac{x}{(5*10) - {(-4)*(-11)}}\)=\(\frac{y}{(-11)3 - (-10)8}\)=\(\frac{1}{8(-4) - (3*5)}\)

\(\frac{x}{(-94)}\)=\(\frac{y}{47}\)=\(\frac{1}{-47}\)

Multiply by 47

\(\frac{x}{-2}\)=\(\frac{y}{1}\)=\(\frac{1}{-1}\)

Hence, the required solution is x = 2, y = -1

Ques. Using the cross-multiplication method, find the value of x and y:
4x – 3y – 6 = 0, 3x + 4y – 17 = 0 (3 marks)

Ans. Two equations are given,

3x + 4y -17 = 0

4x - 3y - 6 = 0

By the method of cross-multiplication,

\(\frac{x}{4(-6) - {(-3)(-17)}}\)=\(\frac{y}{(-17)4 - (-6)3}\)=\(\frac{1}{3(-3) - (4*4)}\)

\(\frac{x}{(-75)}\)=\(\frac{y}{-50}\)=\(\frac{1}{-25}\)

Multiply by 47

\(\frac{x}{3}\)=\(\frac{y}{2}\)=1

Hence, required solution: x = 3, y = 2.

Ques. Solve the linear equations:
ax + by – c² = 0
a²x + b²y – c² = 0 (3 marks)

Ans. The process is as follows:

\(\frac{x}{(-b + b^2)}\) = \(\frac{y}{(-a^2 + a)}\) = \(\frac{1}{(ab^2 + a^2b)}\)

\(\frac{x}{-b (1 - b)}\)=\(\frac{y}{-a (a - 1)}\)=\(\frac{1}{-ab (a - b)}\)

\(\frac{x}{b (1 - b)}\)=\(\frac{y}{a (a - 1)}\)=\(\frac{1}{ab (a - b)}\)

x = \(\frac{bc^2(1 - b)}{ab (a - b)}\)= \(\frac{c^2(1 - b)}{a (a - b)}\)

y = \(\frac{c^2a(a - 1)}{ab (a - b)}\)=\(\frac{c^2(a - 1)}{b (a - b)}\)

Ques. Using the Cross Multiplication Method, solve the linear equations x+y=2 and 2x+3y=4. (4 marks)

Ans. The process is as follows:

x + y = 2

2x + 3y = 4

x + y - 2 = 0

2x + 3y - 4 = 0

In this case,

a₁=1, b₁=1, c₁=-2

a₂=2, b₂=3, c₂=-4

x=\(\frac{b_1c_2 - b_2c_1}{b_2a_1 - b_1a_2}\)

=\(\frac{1(-4) - 3-(-2)}{1*3 - 2*1}\)

=\(\frac{(-4) -(-6)}{3 -2}\)

=\(\frac{1}{2}\)

= 2

y=\(\frac{c_1a_2 - c_2a_1}{b_2a_1 - b_1a_2}\)

=\(\frac{(-2)(2) -(-4)-(1)}{1*3 - 2*1}\)

=\(\frac{(-4) -(-4)}{3 - 2}\)

=0/1

= 0

∴ x=2 and y=0

Ques. What is cross multiplication method? (3 marks)

Ans. The cross multiplication approach is applied to two-variable linear equation solutions. The simplest and most direct way for resolving linear equations in two variables is the cross-multiplication method.

In cases where a linear equation contains two variables, this method is frequently applied.
There are two approaches to solving linear equations: algebraic and graphical.
The three categories of algebraic methods are cross-multiplication, substitution, and elimination.
The denominator on the right is multiplied by the numerator on the left in the cross multiplication method.

Ques. What are the equations for inconsistent solutions? (3 marks)

Ans. The equations for inconsistent solutions are as follows:

a₁b₂ – a₂b₁ = 0
b₁c₂ – b₂c₁ ≠ 0
a₁c₂ – a₂c₁ ≠ 0

Ques. What are the equations for infinitely many solutions? (3 marks)

Ans. The equations for infinitely many solutions are as follows:

a₁b₂ – a₂b₁ = 0
b₁c₂ – b₂c₁ = 0
a₁c₂ – a₂c₁ = 0

Ques. Suppose you are provided the following linear equations: 2x – 3y = 6 and 5x – 6y = 8. Solve them using a cross multiplication method. (3 marks)

Ans. The equation are given as:

2x – 3y = 6 and 5x – 6y = 8
Now compare with a₁x + b₁y = -c₁ and a₂x + b₂y = -c₂
a₁ = 2, b₁ = -3, c¹ = -6
a₂ = 5, b₂ = -6, c² = -8
Using Cross Multiplication Method,
x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(b₂a₁ – b₁a₂₎
Substitute all the values
x/[(-3).(-8) – (-6).(-6)] = y/[(-6).(5) – (-8).(2)] = 1/[(-6).(2) – (-3).(5)]
x/[24 – 36] = y/[-30 + 16] = 1/[-12 + 15]
x/(-12) = y/(-14) = 1/(3)
x/(-12) = 1/3
x = -4
y/(-14) = 1/(3)
y = -14/3
Thus, the value of x = -4 and -14/3

Ques. Suppose you are provided the following linear equations: 3x + 4y = 5 and x + 2y = 6. Solve them using a cross multiplication method. (3 marks)

Ans. The equation are given as:

3x + 4y = 5 and x + 2y = 6
Now compare with a₁x + b₁y = -c₁ and a₂x + b₂y = -c₂
a₁ = 3, b₁ = 4, c_¹= -5
a₂ = 1, b₂ = 2, c_²= -6
Using Cross Multiplication Method,
x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(b₂a₁ – b₁a₂₎
Substitute all the values
x/[(4).(-6) – (2).(-5)] = y/[(-5).(1) – (-6).(3)] = 1/[(2).(3) – (4).(1)]
x/[-24 + 10] = y/[-5 + 18] = 1/[6 – 4]
x/(-14) = y/(13) = 1/(2)
x/(-14) = 1/2
x = -7
y/(13) = 1/(2)
y = 13/2

Cross Multiplication Method: Formula, Derivation & Examples