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)

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● **Simultaneous Linear Equations - Worksheets**

**Worksheet on Simultaneous Linear Equations**

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**8th Grade Math Practice****From Cross-Multiplication Method to HOME PAGE**

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