Tenth Grade Mathematics: Algebraic Solution: Cross-Multiplication Method
While substitution and elimination solve systems through stepwise manipulation, algebra allows us to express the general solution of any two-variable linear system as a direct determinant formula. The Cross-Multiplication Method derives a compact closed-form solution directly from the coefficients of the standard equations. This method is especially powerful in advanced coordinate geometry, vector algebra, and when working with abstract parametric coefficients where standard arithmetic elimination becomes cumbersome.
Derivation of the Cross-Multiplication Formula
Consider the standard general system:
a1*x + b1*y + c1 = 0 --- (Equation 1)
a2*x + b2*y + c2 = 0 --- (Equation 2)
To eliminate y, multiply Equation 1 by b2 and Equation 2 by b1:
b2*(a1*x + b1*y + c1) = 0 ===> b2*a1*x + b2*b1*y + b2*c1 = 0
b1*(a2*x + b2*y + c2) = 0 ===> b1*a2*x + b1*b2*y + b1*c2 = 0
Subtracting the second equation from the first eliminates y:
(a1*b2 - a2*b1)*x + (b2*c1 - b1*c2) = 0
(a1*b2 - a2*b1)*x = -(b2*c1 - b1*c2) = b1*c2 - b2*c1
Provided that a1b2 - a2b1 != 0, we can divide to solve for x:
x = (b1*c2 - b2*c1) / (a1*b2 - a2*b1)
Similarly, multiplying Equation 1 by a2 and Equation 2 by a1 to eliminate x yields:
y = (c1*a2 - c2*a1) / (a1*b2 - a2*b1)
The Mnemonic Cross-Multiplication Diagram
To memorize and apply these formulas effortlessly without memorizing indices, write down the coefficients cyclically starting from b, moving to c, then to a, and returning to b:
x y 1
b1 c1 c1 a1 a1 b1
\ / \ / \ /
/ \ / \ / b2 c2 c2 a2 a2 b2
Taking downward products as positive and upward products as negative gives the unified proportion:
x y 1
------------------- = ------------------- = -------------------
(b1*c2 - b2*c1) (c1*a2 - c2*a1) (a1*b2 - a2*b1)
From this three-way proportion, the variables are read off directly:
x = (b1*c2 - b2*c1) / (a1*b2 - a2*b1)
y = (c1*a2 - c2*a1) / (a1*b2 - a2*b1)
Notice that the denominator (a1b2 - a2b1) is identical for both variables. If this denominator equals zero, the condition a1/a2 = b1/b2 holds, signaling that the system is either inconsistent or dependent.
Detailed Step-by-Step Example
Solve the following system of linear equations using the cross-multiplication method:
2*x + 3*y = 46
3*x + 5*y = 74
Step 1: Write both equations in standard form with zero on the right side:
2*x + 3*y - 46 = 0 ===> a1 = 2, b1 = 3, c1 = -46
3*x + 5*y - 74 = 0 ===> a2 = 3, b2 = 5, c2 = -74
Step 2: Set up the cross-multiplication pattern:
x y 1
3 -46 -46 2 2 3
\ / \ / \ /
/ \ / \ / 5 -74 -74 3 3 5
Step 3: Calculate the three cross-multiplication terms:
Denominator under x:
b1*c2 - b2*c1 = (3)*(-74) - (5)*(-46) = -222 - (-230) = -222 + 230 = 8
Denominator under y:
c1*a2 - c2*a1 = (-46)*(3) - (-74)*(2) = -138 - (-148) = -138 + 148 = 10
Denominator under 1:
a1*b2 - a2*b1 = (2)*(5) - (3)*(3) = 10 - 9 = 1
Step 4: Equate ratios:
x / 8 = y / 10 = 1 / 1
x = 8 / 1 = 8
y = 10 / 1 = 10
The unique solution is x = 8, y = 10.
Chapter Exercises
Exercise 1: Solve the pair of linear equations by cross-multiplication: 2x + y = 5 and 3x + 2*y = 8.
Exercise 2: Solve the parametric system using cross-multiplication: ax + by = a^2 and bx + ay = b^2, where a^2 != b^2.
Exercise 3: Find the values of x and y using the cross-multiplication method: (x/a) + (y/b) = 2 and ax - by = a^2 - b^2.
Solutions and Explanations
Solution 1: Write in standard form: 2x + y - 5 = 0 and 3x + 2y - 8 = 0. Coefficients: a1 = 2, b1 = 1, c1 = -5; a2 = 3, b2 = 2, c2 = -8. Set up the cross-multiplication proportions: x / (b1c2 - b2c1) = y / (c1a2 - c2a1) = 1 / (a1b2 - a2b1). Compute denominators: For x: (1)(-8) - (2)(-5) = -8 + 10 = 2. For y: (-5)(3) - (-8)(2) = -15 + 16 = 1. For 1: (2)(2) - (3)*(1) = 4 - 3 = 1. Thus, x / 2 = y / 1 = 1 / 1. This gives x = 2 and y = 1.
Solution 2: Rewrite in standard form: ax + by - a^2 = 0 and bx + ay - b^2 = 0. Coefficients: a1 = a, b1 = b, c1 = -a^2; a2 = b, b2 = a, c2 = -b^2. Denominator for x: b1c2 - b2c1 = (b)(-b^2) - (a)(-a^2) = -b^3 + a^3 = a^3 - b^3 = (a - b)(a^2 + ab + b^2). Denominator for y: c1a2 - c2a1 = (-a^2)(b) - (-b^2)(a) = -a^2b + ab^2 = -ab(a - b). Denominator for 1: a1b2 - a2b1 = aa - bb = a^2 - b^2 = (a - b)(a + b). Equating x: x = [(a - b)(a^2 + ab + b^2)] / [(a - b)(a + b)] = (a^2 + ab + b^2) / (a + b). Equating y: y = [-ab(a - b)] / [(a - b)(a + b)] = -a*b / (a + b).
Solution 3: Write both equations in standard form. For (x/a) + (y/b) = 2, multiply by ab: bx + ay - 2ab = 0. Second equation: ax - by - (a^2 - b^2) = 0. Coefficients: a1 = b, b1 = a, c1 = -2ab; a2 = a, b2 = -b, c2 = -(a^2 - b^2) = b^2 - a^2. Denominator for x: a(b^2 - a^2) - (-b)(-2ab) = ab^2 - a^3 - 2ab^2 = -a^3 - ab^2 = -a(a^2 + b^2). Denominator for y: (-2ab)(a) - (b^2 - a^2)(b) = -2a^2b - b^3 + a^2b = -a^2b - b^3 = -b(a^2 + b^2). Denominator for 1: b(-b) - aa = -b^2 - a^2 = -(a^2 + b^2). Equating: x = [-a(a^2 + b^2)] / [-(a^2 + b^2)] = a. Equating y: y = [-b*(a^2 + b^2)] / [-(a^2 + b^2)] = b. The solution is x = a, y = b.