Cross-Multiplication Method for Solving Pair of Linear Equations
The cross-multiplication method is a powerful algebraic technique used to solve a pair of linear equations in two variables. Unlike substitution or elimination, which often require multiple rearrangement steps, cross-multiplication provides a direct formula-based approach that yields the values of both unknowns in a single systematic procedure. This method is especially useful when dealing with equations whose coefficients do not lend themselves easily to elimination or when a compact, efficient solution process is desired. For CBSE Class 10 students, mastering the cross-multiplication method is essential because it offers a reliable alternative to other solving techniques, appears frequently in board examinations, and strengthens the understanding of how determinants work in linear algebra. The method derives its name from the cross-wise multiplication of coefficients, forming a memorable pattern that students can apply quickly once they grasp the underlying logic. In this topic, we will explore the formula, its derivation, step-by-step procedure, and work through a comprehensive set of examples to build confidence and fluency.
What is Cross-Multiplication Method?
The cross-multiplication method is a technique for solving a system of two linear equations in two variables by arranging the coefficients in a specific cross-wise pattern and computing ratios. Given the standard form of two linear equations:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
the cross-multiplication method expresses the solution as a ratio involving the coefficients. The method gets its name from the way we 'cross multiply' the coefficients of y with the constants, and the coefficients of x with the constants, in a pattern reminiscent of computing 2x2 determinants.
The key idea is that for a consistent system with a unique solution, the values of x and y can each be written as a fraction whose numerator and denominator are formed by cross-multiplying specific pairs of coefficients. This avoids the need for step-by-step elimination or substitution and delivers the answer directly.
The cross-multiplication method is applicable only when the system has a unique solution, i.e., when a1/a2 is not equal to b1/b2. If a1/a2 = b1/b2, the lines are either parallel (no solution) or coincident (infinitely many solutions), and the cross-multiplication formula will produce a zero denominator, signalling that the method does not yield a unique answer in those cases.
Cross-Multiplication Method for Solving Pair of Linear Equations Formula
Cross-Multiplication Formula:
For the system of equations written in the form:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
x / (b1c2 - b2c1) = y / (c1a2 - c2a1) = 1 / (a1b2 - a2b1)
From this, the solution is:
x = (b1c2 - b2c1) / (a1b2 - a2b1)
y = (c1a2 - c2a1) / (a1b2 - a2b1)
Memory Aid — The Arrow Diagram:
Write the coefficients in three columns: y-coefficients, constants, x-coefficients, y-coefficients. Draw diagonal arrows downward-right (positive) and downward-left (negative) to remember the cross products.
| b | c | a | b | |
|---|---|---|---|---|
| Equation 1 | b1 | c1 | a1 | b1 |
| Equation 2 | b2 | c2 | a2 | b2 |
Then: x is found from the first pair of columns, y from the second pair, and the denominator from the third pair.
Derivation and Proof
The cross-multiplication formula can be derived from the general system using the elimination method.
Step 1: Start with the two equations:
a1x + b1y + c1 = 0 ... (i)
a2x + b2y + c2 = 0 ... (ii)
Step 2: Multiply equation (i) by b2 and equation (ii) by b1:
a1b2x + b1b2y + c1b2 = 0 ... (iii)
a2b1x + b1b2y + c2b1 = 0 ... (iv)
Step 3: Subtract equation (iv) from equation (iii):
(a1b2 - a2b1)x + (c1b2 - c2b1) = 0
Step 4: Solve for x:
x = (b1c2 - b2c1) / (a1b2 - a2b1)
(Note: we rearranged signs to match the standard formula notation.)
Step 5: Similarly, multiply equation (i) by a2 and equation (ii) by a1, then subtract to eliminate x:
(a2b1 - a1b2)y + (a2c1 - a1c2) = 0
Step 6: Solve for y:
y = (c1a2 - c2a1) / (a1b2 - a2b1)
Step 7: Both expressions share the same denominator (a1b2 - a2b1), which must be non-zero for a unique solution. Writing them as a chain of equal ratios gives the cross-multiplication formula.
The denominator a1b2 - a2b1 is the determinant of the coefficient matrix. When this determinant is zero, the system either has no solution or infinitely many solutions.
Methods
Step-by-Step Procedure for Cross-Multiplication:
Step 1: Write both equations in the standard form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Ensure that all terms are on one side with zero on the other. Be extremely careful about signs when rearranging.
Step 2: Identify the six coefficients: a1, b1, c1, a2, b2, c2.
Step 3: Set up the cross-multiplication scheme. Write the coefficients in the pattern: b, c, a, b (repeating the b column). Compute three cross products:
- For x (numerator): b1c2 - b2c1
- For y (numerator): c1a2 - c2a1
- Denominator: a1b2 - a2b1
Step 4: Compute x and y by dividing the respective numerators by the denominator.
Step 5: Verify by substituting both values into both original equations to ensure both are satisfied.
Important Precautions:
- Always rearrange to the form ax + by + c = 0, not ax + by = c. The constant must be moved to the left side.
- The sign of c matters critically. If the original equation is 2x + 3y = 7, rewrite as 2x + 3y - 7 = 0, making c = -7.
- If the denominator (a1b2 - a2b1) equals zero, the method does not apply — the system is either inconsistent or has infinitely many solutions.
Solved Examples
Example 1: Basic Cross-Multiplication with Integer Coefficients
Problem: Solve using cross-multiplication: 2x + 3y = 11 and 3x - y = 5.
Solution:
Step 1: Rewrite in standard form:
2x + 3y - 11 = 0 ... (i)
3x - y - 5 = 0 ... (ii)
Step 2: Identify coefficients:
a1 = 2, b1 = 3, c1 = -11
a2 = 3, b2 = -1, c2 = -5
Step 3: Apply cross-multiplication formula:
x / (b1c2 - b2c1) = y / (c1a2 - c2a1) = 1 / (a1b2 - a2b1)
Step 4: Compute each cross product:
b1c2 - b2c1 = (3)(-5) - (-1)(-11) = -15 - 11 = -26
c1a2 - c2a1 = (-11)(3) - (-5)(2) = -33 + 10 = -23
a1b2 - a2b1 = (2)(-1) - (3)(3) = -2 - 9 = -11
Step 5: Find x and y:
x = -26 / -11 = 26/11
y = -23 / -11 = 23/11
Step 6: Verification:
2(26/11) + 3(23/11) = 52/11 + 69/11 = 121/11 = 11 ✓
3(26/11) - 23/11 = 78/11 - 23/11 = 55/11 = 5 ✓
Answer: x = 26/11 and y = 23/11.
Example 2: Cross-Multiplication with Simple Integer Solutions
Problem: Solve by cross-multiplication: x + 2y = 1 and 2x - 3y = 12.
Solution:
Step 1: Standard form:
x + 2y - 1 = 0 ... (i)
2x - 3y - 12 = 0 ... (ii)
Step 2: a1 = 1, b1 = 2, c1 = -1, a2 = 2, b2 = -3, c2 = -12
Step 3: Cross products:
b1c2 - b2c1 = (2)(-12) - (-3)(-1) = -24 - 3 = -27
c1a2 - c2a1 = (-1)(2) - (-12)(1) = -2 + 12 = 10
a1b2 - a2b1 = (1)(-3) - (2)(2) = -3 - 4 = -7
Step 4: x = -27 / -7 = 27/7, y = 10 / -7 = -10/7
Step 5: Verification:
27/7 + 2(-10/7) = 27/7 - 20/7 = 7/7 = 1 ✓
2(27/7) - 3(-10/7) = 54/7 + 30/7 = 84/7 = 12 ✓
Answer: x = 27/7 and y = -10/7.
Example 3: Cross-Multiplication with Negative Coefficients
Problem: Solve using cross-multiplication: 3x - 5y + 1 = 0 and x - 2y + 1 = 0.
Solution:
Step 1: Equations are already in standard form:
3x - 5y + 1 = 0 ... (i)
x - 2y + 1 = 0 ... (ii)
Step 2: a1 = 3, b1 = -5, c1 = 1, a2 = 1, b2 = -2, c2 = 1
Step 3: Cross products:
b1c2 - b2c1 = (-5)(1) - (-2)(1) = -5 + 2 = -3
c1a2 - c2a1 = (1)(1) - (1)(3) = 1 - 3 = -2
a1b2 - a2b1 = (3)(-2) - (1)(-5) = -6 + 5 = -1
Step 4: x = -3 / -1 = 3, y = -2 / -1 = 2
Step 5: Verification:
3(3) - 5(2) + 1 = 9 - 10 + 1 = 0 ✓
3 - 2(2) + 1 = 3 - 4 + 1 = 0 ✓
Answer: x = 3 and y = 2.
Example 4: Cross-Multiplication Yielding Fractional Answers
Problem: Solve by cross-multiplication: 5x + 4y = 14 and 4x + 5y = 13.
Solution:
Step 1: Standard form:
5x + 4y - 14 = 0 ... (i)
4x + 5y - 13 = 0 ... (ii)
Step 2: a1 = 5, b1 = 4, c1 = -14, a2 = 4, b2 = 5, c2 = -13
Step 3: Cross products:
b1c2 - b2c1 = (4)(-13) - (5)(-14) = -52 + 70 = 18
c1a2 - c2a1 = (-14)(4) - (-13)(5) = -56 + 65 = 9
a1b2 - a2b1 = (5)(5) - (4)(4) = 25 - 16 = 9
Step 4: x = 18/9 = 2, y = 9/9 = 1
Step 5: Verification:
5(2) + 4(1) = 10 + 4 = 14 ✓
4(2) + 5(1) = 8 + 5 = 13 ✓
Answer: x = 2 and y = 1.
Example 5: Cross-Multiplication with Larger Coefficients
Problem: Solve using cross-multiplication: 7x - 2y = 17 and 5x + 3y = 14.
Solution:
Step 1: Standard form:
7x - 2y - 17 = 0 ... (i)
5x + 3y - 14 = 0 ... (ii)
Step 2: a1 = 7, b1 = -2, c1 = -17, a2 = 5, b2 = 3, c2 = -14
Step 3: Cross products:
b1c2 - b2c1 = (-2)(-14) - (3)(-17) = 28 + 51 = 79
c1a2 - c2a1 = (-17)(5) - (-14)(7) = -85 + 98 = 13
a1b2 - a2b1 = (7)(3) - (5)(-2) = 21 + 10 = 31
Step 4: x = 79/31, y = 13/31
Step 5: Verification:
7(79/31) - 2(13/31) = 553/31 - 26/31 = 527/31 = 17 ✓
5(79/31) + 3(13/31) = 395/31 + 39/31 = 434/31 = 14 ✓
Answer: x = 79/31 and y = 13/31.
Example 6: Cross-Multiplication with One Zero Coefficient
Problem: Solve by cross-multiplication: 4x + 3y = 18 and 3x = 12.
Solution:
Step 1: Standard form:
4x + 3y - 18 = 0 ... (i)
3x + 0y - 12 = 0 ... (ii)
Step 2: a1 = 4, b1 = 3, c1 = -18, a2 = 3, b2 = 0, c2 = -12
Step 3: Cross products:
b1c2 - b2c1 = (3)(-12) - (0)(-18) = -36 - 0 = -36
c1a2 - c2a1 = (-18)(3) - (-12)(4) = -54 + 48 = -6
a1b2 - a2b1 = (4)(0) - (3)(3) = 0 - 9 = -9
Step 4: x = -36 / -9 = 4, y = -6 / -9 = 2/3
Step 5: Verification:
4(4) + 3(2/3) = 16 + 2 = 18 ✓
3(4) = 12 ✓
Answer: x = 4 and y = 2/3.
Example 7: Cross-Multiplication Applied to a Word Problem Setup
Problem: The sum of two numbers is 35 and their difference is 7. Find the numbers using cross-multiplication.
Solution:
Step 1: Let the two numbers be x and y. Form the equations:
x + y = 35 → x + y - 35 = 0 ... (i)
x - y = 7 → x - y - 7 = 0 ... (ii)
Step 2: a1 = 1, b1 = 1, c1 = -35, a2 = 1, b2 = -1, c2 = -7
Step 3: Cross products:
b1c2 - b2c1 = (1)(-7) - (-1)(-35) = -7 - 35 = -42
c1a2 - c2a1 = (-35)(1) - (-7)(1) = -35 + 7 = -28
a1b2 - a2b1 = (1)(-1) - (1)(1) = -1 - 1 = -2
Step 4: x = -42 / -2 = 21, y = -28 / -2 = 14
Step 5: Verification: 21 + 14 = 35 ✓ and 21 - 14 = 7 ✓
Answer: The two numbers are 21 and 14.
Example 8: Cross-Multiplication Where Constants Need Rearranging
Problem: Solve by cross-multiplication: 3x = 8 - 2y and 4y = 6 - 5x.
Solution:
Step 1: Rearrange to standard form:
3x + 2y - 8 = 0 ... (i)
5x + 4y - 6 = 0 ... (ii)
Step 2: a1 = 3, b1 = 2, c1 = -8, a2 = 5, b2 = 4, c2 = -6
Step 3: Cross products:
b1c2 - b2c1 = (2)(-6) - (4)(-8) = -12 + 32 = 20
c1a2 - c2a1 = (-8)(5) - (-6)(3) = -40 + 18 = -22
a1b2 - a2b1 = (3)(4) - (5)(2) = 12 - 10 = 2
Step 4: x = 20/2 = 10, y = -22/2 = -11
Step 5: Verification:
3(10) = 30, and 8 - 2(-11) = 8 + 22 = 30 ✓
4(-11) = -44, and 6 - 5(10) = 6 - 50 = -44 ✓
Answer: x = 10 and y = -11.
Example 9: NCERT-Style Problem Using Cross-Multiplication
Problem: Solve by cross-multiplication: 2x + y = 5 and 3x + 2y = 8.
Solution:
Step 1: Standard form:
2x + y - 5 = 0 ... (i)
3x + 2y - 8 = 0 ... (ii)
Step 2: a1 = 2, b1 = 1, c1 = -5, a2 = 3, b2 = 2, c2 = -8
Step 3: Cross products:
b1c2 - b2c1 = (1)(-8) - (2)(-5) = -8 + 10 = 2
c1a2 - c2a1 = (-5)(3) - (-8)(2) = -15 + 16 = 1
a1b2 - a2b1 = (2)(2) - (3)(1) = 4 - 3 = 1
Step 4: x = 2/1 = 2, y = 1/1 = 1
Step 5: Verification:
2(2) + 1 = 5 ✓
3(2) + 2(1) = 8 ✓
Answer: x = 2 and y = 1.
Example 10: Cross-Multiplication with All Negative Coefficients
Problem: Solve using cross-multiplication: -2x - 3y + 13 = 0 and -x + 4y - 11 = 0.
Solution:
Step 1: Equations are in standard form:
-2x - 3y + 13 = 0 ... (i)
-x + 4y - 11 = 0 ... (ii)
Step 2: a1 = -2, b1 = -3, c1 = 13, a2 = -1, b2 = 4, c2 = -11
Step 3: Cross products:
b1c2 - b2c1 = (-3)(-11) - (4)(13) = 33 - 52 = -19
c1a2 - c2a1 = (13)(-1) - (-11)(-2) = -13 - 22 = -35
a1b2 - a2b1 = (-2)(4) - (-1)(-3) = -8 - 3 = -11
Step 4: x = -19 / -11 = 19/11, y = -35 / -11 = 35/11
Step 5: Verification:
-2(19/11) - 3(35/11) + 13 = -38/11 - 105/11 + 143/11 = 0 ✓
-(19/11) + 4(35/11) - 11 = -19/11 + 140/11 - 121/11 = 0 ✓
Answer: x = 19/11 and y = 35/11.
Real-World Applications
The cross-multiplication method has several practical and theoretical applications:
Quick Solutions in Competitive Exams: In timed examinations such as Olympiads and entrance tests, the cross-multiplication method provides a fast, formula-based route to solve linear systems without the trial-and-error of elimination or substitution.
Foundation for Determinants: The cross-multiplication formula is actually an application of Cramer's Rule using 2x2 determinants. Students who master this method in Class 10 find it much easier to learn matrix algebra and determinants in Classes 11 and 12.
Checking Consistency: By examining whether the denominator (a1b2 - a2b1) is zero, students can quickly determine if a system has a unique solution, no solution, or infinitely many solutions, without needing to solve the full equations.
Business and Economics: Systems of linear equations model supply-demand relationships, cost-revenue analysis, and resource allocation problems. The cross-multiplication method provides an efficient way to solve such systems when the equations are already identified.
Geometry: Finding the point of intersection of two straight lines involves solving a pair of linear equations. Cross-multiplication gives the intersection point directly from the line equations.
Key Points to Remember
- The cross-multiplication method solves a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 using: x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1).
- Always rewrite both equations in the form ax + by + c = 0 before applying the formula.
- The constant term c includes its sign. For example, 2x + 3y = 7 becomes 2x + 3y - 7 = 0, so c = -7.
- The method works only when a1b2 - a2b1 is not equal to 0 (unique solution exists).
- If a1b2 - a2b1 = 0, the system is either inconsistent (parallel lines) or dependent (coincident lines).
- The arrow diagram (b, c, a, b pattern) is a useful memory aid for remembering the cross products.
- Always verify the solution by substituting back into both original equations.
- The denominator a1b2 - a2b1 is the same as the determinant of the coefficient matrix.
- Cross-multiplication is a direct precursor to Cramer's Rule studied in higher classes.
Practice Problems
- Solve by cross-multiplication: x + 3y = 7 and 2x - y = 7.
- Use cross-multiplication to solve: 3x - 4y = 1 and 2x + 3y = 12.
- Solve by cross-multiplication: 5x + 2y = 16 and 7x + 4y = 24.
- Apply cross-multiplication to: 6x - 5y = 1 and 3x + 2y = 14.
- Solve using cross-multiplication: x/2 + y/3 = 4 and x/3 + y/2 = 11/3.
- The sum of the numerator and denominator of a fraction is 8. If 1 is added to both, the fraction becomes 3/5. Find the fraction using cross-multiplication.
Frequently Asked Questions
Q1. What is the cross-multiplication method?
The cross-multiplication method is a formula-based technique for solving a pair of linear equations in two variables. It uses a specific cross-wise pattern of multiplying coefficients to directly compute the values of x and y without step-by-step elimination or substitution.
Q2. When can the cross-multiplication method be used?
The cross-multiplication method can be used whenever the pair of linear equations has a unique solution, i.e., when the denominator a₁b₂ - a₂b₁ ≠ 0. If this denominator equals zero, the system is either inconsistent or dependent, and the method does not apply.
Q3. Why must equations be written in the form ax + by + c = 0?
The cross-multiplication formula is derived assuming both equations are in the form ax + by + c = 0. If the equations are in another form (like ax + by = k), the signs of the coefficients will be wrong, leading to an incorrect answer. Moving all terms to the left side ensures the formula works correctly.
Q4. How is cross-multiplication different from the elimination method?
Elimination involves manipulating the equations to cancel one variable, then solving for the other. Cross-multiplication uses a direct formula to find both variables simultaneously, without any intermediate steps of adding or subtracting equations. Cross-multiplication is often faster but requires careful handling of signs.
Q5. What happens when a₁b₂ - a₂b₁ = 0 in cross-multiplication?
When this denominator is zero, it means the two lines are either parallel (no solution) or coincident (infinitely many solutions). In either case, the cross-multiplication formula cannot give a unique answer, and we must use other methods to determine the type of system.
Q6. Is cross-multiplication important for CBSE Board exams?
Yes, cross-multiplication is a commonly asked method in CBSE Class 10 board exams. Questions specifically asking students to 'solve by cross-multiplication' appear regularly. Understanding this method also helps in solving problems faster during examinations.
Q7. How do I remember the cross-multiplication formula?
Use the arrow diagram: write coefficients in the order b, c, a, b (repeating b). Draw diagonal arrows. The products along downward-right arrows minus those along downward-left arrows give the numerators for x, y, and the common denominator.
Q8. Can cross-multiplication be used for three equations?
The standard cross-multiplication method as taught in Class 10 is specifically for two equations in two unknowns. For three equations in three unknowns, Cramer's Rule (using 3×3 determinants) is the generalized version, which is studied in Classes 11–12.
Q9. What is the relationship between cross-multiplication and determinants?
The cross-multiplication formula is essentially Cramer's Rule applied to a 2×2 system. The denominator a₁b₂ - a₂b₁ is the determinant of the coefficient matrix, and the numerators are determinants formed by replacing one column of the coefficient matrix with the constants column.
Q10. Do I need to verify my answer after cross-multiplication?
Yes, verification is always recommended. Substitute both values of x and y back into both original equations to confirm they are satisfied. This catches sign errors, which are the most common mistakes in cross-multiplication.
Related Topics
- Elimination Method
- Substitution Method
- Pair of Linear Equations in Two Variables
- Consistency of Linear Equations
- Graphical Method for Linear Equations
- Word Problems on Pair of Linear Equations
- Word Problems Solved Graphically
- Reducing Equations to Linear Form
- Conditions for Solvability
- Linear Equations in Real Life
