# Cross Multiplication Method

Cross multiplication method is one of the basic methods in mathematics that is used to solve the linear equations in two variables. It is one of the easiest to solve a pair of linear equations in two variables.Â

Suppose we have a pair of linear equations in two variables, i.e. a1x + b1y = -c1 and a2x + b2y = -c2 then we can directly get their solution by using the cross multiplication method. This method is applied only when the condition,

b2a1 – b1a2 â‰  0

## Definition of Cross Multiplication Method

Cross-multiplication is the fundamental technique that is used to solve linear equations in two variables. If applied correctly it is one of the fastest methods to solve linear equations in two variables. For example, if we have given pair of linear equations in two variables,

(a)1x + (b)1y + (c)1 = 0

(a)2x + (b)2y + (c)2 = 0

Then using the cross-multiplication method, we easily get the values x and y by solving the fraction.

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### Cross Multiplication Formula

Cross Multiplication Formula is the formula that is used to solve linear equations with two variables. If the given pair of linear equations is,

• (a)1x + (b)1y + (c)1 = 0
• (a)2x + (b)2y + (c)2 = 0

The required cross-multiplication formula is,

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(b2a1 – Â b1a2)

## Derivation of Cross Multiplication Method

We can easily derive the cross multiplication method formula for a pair of linear equations in two variables by simply solving the linear equation by making their coefficients equal. This can be understood as,

Let’s take two linear equations as

• (a)1x + (b)1y + (c)1 = 0…(i)
• (a)2x + (b)2y + (c)2 = 0…(ii)

For solving the above pair of linear equations we make the coefficient of y in both equations equal.

Multiplying b2 to eq(i) and b1 to eq(ii)

(b2)(a)1x + (b2)(b)1y + (b2)(c)1 = 0…(iii)

(b1)(a)2x + (b1)(b)2y + (b1)(c)2 = 0…(iv)

Subtracting equations (iii) and (iv) we get

(b2a1 – b1a2)x + (b2c1 – b1c2) = 0

Solving for x we get,

x = (b1c2 – b2c1)/(b2a1 – b1a2)…(a)

where, (b2a1 – b1a2) â‰  0

Similarly solving eq (i) and eq (ii) for y we get,

y = (c1a2 – c2a1)/(b2a1 – b1a2)…(b)

where, (b2a1 – b1a2) â‰  0

Combining eq.(a) and eq.(b) we get,

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(b2a1 – Â b1a2)

This is the required cross-multiplication formula.

We can easily write this formula by using the technique discussed in the image below,

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## Solving Linear Equations by Cross Multiplication Method

The solution to a pair of linear equations with two variables can be easily achieved using the Cross Multiplication method. We use the Cross Multiplication formula for finding the solution to the pair of linear equations with two variables.

The Cross-Multiplication formula is,

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(b2a1 – Â b1a2)

Follow the steps discussed below to get the solution to the linear equations,

• 2x + 3y = 5
• x + y = 3

Step 1: Arrange the linear equation in the general form as, ax + by + c = 0

2x + 3y – 5 = 0

x + y – 3 = 0

Step 2: Compare with a1x + b1y = -c1 and a2x + b2y = -c2 and find all the coefficients.

a1 = 2, b1 = 3, c1 = -5

a2 = 1, b2 = 1, c2 = -3

Step 3: Use the Cross-multiplication formula and substitute all the values.

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(b2a1 – Â b1a2)

Substituting all the values we get,

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

Step 4: Simplify the fractions

x/ (-9+5) = y/(-5+6) = 1/(2-3)

x/(-4) = y/(1) = 1/(-1)

Step 5: Compare the first and the third fraction to get x

x/(-4) = 1/(-1)

x = 4

Step 6: Compare the second and the third fraction to get x

y/(1) = 1/(-1)

y = -1

## Unique Solution by Cross Multiplication Method

We get a unique solution to the pair of linear equations if the given pair of lines are consistent. For any two lines a1x + b1y = -c1 and a2x + b2y = -c2 to be consistent the required condition is,

a1/a2 â‰  b1/b2

This situation can also be represented as,

a1b1 â‰  a2b2

a1b1 – a2b2 = 0

In case of a unique solution, the line for two equations will intersect at a unique point which is the solution of the two equations.

For example, Check whether 2x + 3y = 5 and 3x + y = 7 yields a unique solution or not.

Solution:

Given Equation,

• 2x + 3y = 5
• 3x + y = 7

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

a1 = 2, b1 = 3

a2 = 3, b2 = 1

Using the conditionÂ

a1/a2 â‰  b1/b2

2/3 â‰  3/1

As the condition holds true the given equations give the unique solution.

### Inconsistent Solution

We can have an Inconsistent solution to the pair of linear equations if the given pair of lines are parallel but their y-intercepts are not equal. For any two lines a1x + b1y = -c1 and a2x + b2y = -c2 to have Inconsistent solution the required condition is,

a1/a2 = b1/b2 â‰  c1/c2

This can be further expanded as,

• a1b2 – a2b1 = 0
• b1c2 – b2c1 â‰  0
• a1c2 – a2c1 â‰  0

If the line follows the above condition then it has an Inconsistent Solution. In case of an inconsistent solution, the graph of two equations will be parallel, and hence no solution is derived.

### Infinitely Many Solutions

We can have Infinitely Many solutions to the pair of linear equations if the given pair of lines are parallel but their y-intercepts are also multiple of each other. For any two lines a1x + b1y = -c1 and a2x + b2y = -c2 to have Inconsistent solution the required condition is,

a1/a2 = b1/b2 = c1/c2

This can be further expanded as,

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

If the line follows the above condition then it has Infinitely Many Solutions. In the case of Infinitely many solutions, the two graphs will be overlapping on each other thus giving infinitely many solutions.

## Solved Examples on Cross Multiplication Method

Example 1: Solve the following linear equations using the cross-multiplication method.

• 2x – 3y = 6
• 5x – 6y = 8

Solution:

Given equations are,

2x – 3y = 6 and 5x – 6y = 8

Comparing 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 – Â b1a2)

Substituting 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

Example 2: Solve the following linear equations using the cross-multiplication method.

• 3x + 4y = 5
• x + 2y = 6

Solution:

Given equations are,

3x + 4y = 5 and x + 2y = 6

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

a1 = 3, b1 = 4, c1 = -5

a2 = 1, b2 = 2, c2 = -6

Using Cross Multiplication Method,

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(b2a1 – Â b1a2)

Substituting 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

Thus, the value of x = -7 and 13/2

## FAQs on Cross-Multiplication Method

### Q1: What is a Cross-Multiplication Method?

The cross-multiplication method is a method for finding the solution to the linear equations in two variables in a quick and easy manner. In this method, the numerator of one fraction is multiplied by the denominator of the other and the denominator of the first term is multiplied by the numerator of another term.

### Q2: How to solve using the Cross Multiplication Method?

We can easily solve the linear equation in two variables by using the cross-multiplication formula that is, the solution of two linear equations a1x + b1y = -c1 and a2x + b2y = -c2 can be given using the formula,

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(b2a1 – Â b1a2)

### Q2: What is the Condition to Get a Unique Solution?

We get a unique solution to the pair of linear equations if the lines are consistent and the condition for lines to be consistent is,

a1/a2 â‰  b1/b2

### Q4: What is the Mathematical Rule of Three?

The mathematical rule of three is a method which uses proportions to find the solution of the equations. Â The cross-multiplication method that uses the mathematical rule of three is used for finding the solution to linear equations in two variables.