1. Introduction
The cross multiplication method is a fast and systematic way to solve a pair of linear equations in two variables. It works when both equations are written in the standard form. This method directly gives the values of \(x\) and \(y\) using a simple formula based on cross products.
This method is especially useful when elimination or substitution becomes lengthy.
2. Standard Form Required
The cross multiplication method can be used only when equations are written in the standard form:
\(a_1x + b_1y + c_1 = 0\)
\(a_2x + b_2y + c_2 = 0\)
If the equations are not in this form, rearrange them first.
2.1. Examples of Standard Form
- \(2x + 3y - 7 = 0\)
- \(4x - 5y + 9 = 0\)
- \(3a + 2b - 6 = 0\)
3. Cross Multiplication Formula
For the system:
\(a_1x + b_1y + c_1 = 0\)
\(a_2x + b_2y + c_2 = 0\)
The formula is:
\[ \dfrac{x}{b_1c_2 - b_2c_1} = \dfrac{y}{c_1a_2 - c_2a_1} = \dfrac{1}{a_1b_2 - a_2b_1} \]
Each ratio gives the values of \(x\) and \(y\) after cross multiplication.
3.1. Understanding the Formula
The formula works by pairing terms across the equations in a cross pattern. Each variable’s numerator includes a cross difference of coefficients.
3.1.1. When the Denominator is Zero
If \(a_1b_2 - a_2b_1 = 0\), the method gives:
- No Solution → if other cross-differences don't match
- Infinite Solutions → if all cross-differences match
4. Steps to Solve Using Cross Multiplication
You can solve any pair of linear equations using these steps:
4.1. Step 1: Write Both Equations in Standard Form
Rearrange both equations to match:
\(a_1x + b_1y + c_1 = 0\)
\(a_2x + b_2y + c_2 = 0\)
4.2. Step 2: Identify Coefficients
Identify values of \(a_1, b_1, c_1\) and \(a_2, b_2, c_2\).
4.3. Step 3: Apply Cross Multiplication Formula
Use the formula:
\[ \dfrac{x}{b_1c_2 - b_2c_1} = \dfrac{y}{c_1a_2 - c_2a_1} = \dfrac{1}{a_1b_2 - a_2b_1} \]
4.4. Step 4: Solve for x and y
Use proportional values to find \(x\) and \(y\).
5. Worked Examples
Let’s solve some equations step by step using the formula.
5.1. Example 1
Solve:
\(2x + 3y = 13\)
\(3x - y = 5\)
Convert to standard form:
\(2x + 3y - 13 = 0\)
\(3x - y - 5 = 0\)
Identify coefficients:
- \(a_1 = 2, b_1 = 3, c_1 = -13\)
- \(a_2 = 3, b_2 = -1, c_2 = -5\)
Apply formula:
\(x = \dfrac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} = \dfrac{3(-5) - (-1)(-13)}{2(-1) - 3(3)} = \dfrac{-15 - 13}{-2 - 9} = \dfrac{-28}{-11} = \dfrac{28}{11}\)
\(y = \dfrac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} = \dfrac{(-13)(3) - (-5)(2)}{-11} = \dfrac{-39 +10}{-11} = \dfrac{-29}{-11} = \dfrac{29}{11}\)
Solution: \((x, y) = (\dfrac{28}{11}, \dfrac{29}{11})\)
5.2. Example 2
Solve:
\(3x + 2y = 11\)
\(6x + 4y = 22\)
These equations are multiples of each other, so:
\(a_1b_2 - a_2b_1 = 3 \cdot 4 - 6 \cdot 2 = 12 - 12 = 0\)
Since all cross differences also become zero, the equations are coincident.
Result: Infinitely many solutions.
6. Common Mistakes
- Not converting equations into standard form.
- Sign errors in cross multiplication.
- Incorrect placement of coefficients.
- Dividing by zero without checking special cases.
7. Quick Practice
Solve the following using the cross multiplication method:
- \(x + y = 7\), \(2x - y = 1\)
- \(3x - 2y = 5\), \(4x + y = 11\)
- \(2a + 3b = 9\), \(4a + 6b = 18\)
8. Summary
- The cross multiplication method solves linear equations quickly using a direct formula.
- Equations must be written in standard form.
- The solution is found by evaluating cross differences.
- If the denominator becomes zero, check for parallel or coincident lines.