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:
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:
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
Solvability of Linear Simultaneous Equations
Word Problems on Simultaneous Linear 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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