Cross-Multiplication Method



Here we will discuss about simultaneous linear equations by using cross-multiplication method.



General form of a linear equation in two unknown quantities:
ax + by + c = 0, (a, b ≠ 0)

Two such equations can be written as:
a1x + b1y + c1 = 0 ----------- (i)
a2x + b2y + c2 = 0 ----------- (ii)

Let us solve the two equations by the method of elimination, multiplying both sides of equation (i) by a2 and both sides of equation (ii) by a1, we get:
a1a2x + b1a2y + c1a2 = 0
a1 a2x + a1b2y + a1c2 = 0
Subtracting, b1a2y - a1b2y + c1a2 - c2a1 = 0
or, y(b1 a2 - b2a1) = c2a1 - c1a2
Therefore, y = (c2a1 - c1a2)/(b1a2 - b2a1) = (c1a2 - c2a1)/(a1b2 - a2b1) where (a1b2 - a2b1) ≠ 0
Therefore, y/(c1a2 - c2a1) = 1/(a1b2 - a2b1), ------------- (iii)

Again, multiplying both sides of (i) and (ii) by b2 and b1 respectively, we get;
a1b2x + b1b2y + b2c1 = 0
a2b1x + b1b2y + b1c2 = 0
Subtracting, a1b2x - a2b1x + b2c1 - b1c2 = 0
or, x(a1b2 - a2b1) = (b1c2 - b2c1)
or, x = (b1c2 - b2c1)/(a1b2 - a2b1)
Therefore, x/(b1c2 - b2c1) = 1/(a1b2 - a2b1) where (a1b2 - a2b1) ≠ 0 -------------- (iv)

From equations (iii) and (iv), we get:
x/(b1c2 - b2c1) = y/(c1a2) - c2a1 = 1/(a1b2 - a2b1) where (a1b2 - a2b1) ≠ 0

This relation informs us how the solution of the simultaneous equations, co-efficient x, y and the constant terms in the equations are inter-related, we can take this relation as a formula and use it to solve any two simultaneous equations. Avoiding the general steps of elimination, we can solve the two simultaneous equations directly.

So, the formula for cross-multiplication and its use in solving two simultaneous equations can be presented as:
If (a1b2 - a2b1) ≠ 0 from the two simultaneous linear equations
a1x + b1y + c1 = 0 ----------- (i)
a2x + b2y + c2 = 0 ----------- (ii)

we get, by the cross-multiplication method:
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1) ---------- (A)
That means, x = (b1c2 - b2c1)/(a1b2 - a2b1)
y = (c1a2 - c2a1)/(a1b2 - a2b1)

Note:
If the value of x or y is zero, that is, (b1c2 - b2c1) = 0 or (c1a2 - c2a1) = 0, it is not proper to express in the formula for cross- multiplication, because the denominator of a fraction can never be 0.

From the two simultaneous equations, it appears that the formation of relation (A) by cross-multiplication is the most important concept.

At first, express the co-efficient of the two equations as in the following form:

cross-multiplication method



Now multiply the co-efficient according to the arrow heads and subtract the upward product from the downward product. Place the three differences under x, y and 1 respectively forming three fractions; connect them by two signs of equality.



Worked-out examples on simultaneous linear equations by using cross-multiplication method:

1. Solve the two variables linear equation:
8x + 5y = 11
3x – 4y = 10

Solution:

On transposition, we get
8x + 5y – 11 = 0
3x – 4y – 10 = 0

Writing the co-efficient in the following way, we get:

cross-multiplication, cross multiplication method



Note: The above presentation is not compulsory for solving.
By cross-multiplication method:
x/(5) (-10) – (-4) (-11) = y/(-11) (3) – (-10) (8) = 1/(8) (-4) – (3) (5)
or, x/-50 – 44 = y/-33 + 80 = 1/-32 – 15
or, x/-94 = y/47 = 1/-47
or, x/-2 = y/1 = 1/-1 [multiplying by 47]
or, x = -2/-1 = 2 and y = 1/-1 = -1
Therefore, required solution is x = 2, y = -1


2. Find the value of x and y by using the using cross-multiplication method:
3x + 4y – 17 = 0
4x – 3y – 6 = 0

Solution:

Two given equations are:
3x + 4y – 17 = 0
4x – 3y – 6 = 0

By cross-multiplication, we get:
x/(4) (-6) – (-3) (-17) = y/(-17) (4) – (-6) (3) = 1/(3) (-3) – (4) (4)
or, x/(-24 – 51) = y/(-68 + 18) = 1/(-9 – 16)
or, x/-75 = y/-50 = 1/-25
or, x/3 = y/2 = 1 (multiplying by -25)
or, x = 3, y = 2
Therefore, required solution: x = 3, y = 2.


3. Solve the system of linear equations:
ax + by – c2 = 0
a2x + b2y – c2 = 0

Solution:

x/(-b + b2) = y/(- a2 + a) = c2/(ab2 - a2b)
or, x/-b(1 - b) = y/- a(a - 1) = c2/-ab(a - b)
or, x/b(1 - b) = y/a(a - 1) = c2/ab(a - b)
or, x = bc2(1 – b)/ab(a – b) = c2(1 – b)/a(a – b) and y = c2a(a – 1)/ab(a – b) = c2(a – 1)/b(a – b)

Hence the required solution is:
x = c2(1 – b)/a(a – b)
y = c2a(a – 1)/b(a – b)



Simultaneous Linear Equations

  • Simultaneous Linear Equations
  • Comparison Method
  • Elimination Method
  • Substitution Method
  • Cross-Multiplication Method
  • Solvability of Linear Simultaneous Equations
  • Pairs of Equations
  • Word Problems on Simultaneous Linear Equations
  • Practice Test on Word Problems Involving Simultaneous
    Linear Equations


  • Simultaneous Linear Equations - Worksheets
  • Worksheet on Simultaneous Linear Equations
  • Worksheet on Problems on Simultaneous Linear
    Equations


  • 8th Grade Math Practice

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