## 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:

a_{1}x + b_{1}y + c_{1} = 0 ----------- (i)

a_{2}x + b_{2}y + c_{2} = 0 ----------- (ii)

Let us solve the two equations by the method of elimination, multiplying both sides of equation (i) by a_{2} and both sides of equation (ii) by a_{1}, we get:

a_{1}a_{2}x + b_{1}a_{2}y + c_{1}a_{2} = 0

a_{1} a_{2}x + a_{1}b_{2}y + a_{1}c_{2} = 0

Subtracting, b_{1}a_{2}y - a_{1}b_{2}y + c_{1}a_{2} - c_{2}a_{1} = 0

or, y(b_{1} a_{2} - b_{2}a_{1}) = c_{2}a_{1} - c_{1}a_{2}

Therefore, y = (c_{2}a_{1} - c_{1}a_{2})/(b_{1}a_{2} - b_{2}a_{1}) = (c_{1}a_{2} - c_{2}a_{1})/(a_{1}b_{2} - a_{2}b_{1}) where (a_{1}b_{2} - a_{2}b_{1}) ≠ 0

Therefore, y/(c_{1}a_{2} - c_{2}a_{1}) = 1/(a_{1}b_{2} - a_{2}b_{1}), ------------- (iii)

Again, multiplying both sides of (i) and (ii) by b_{2} and b_{1} respectively, we get;

a_{1}b_{2}x + b_{1}b_{2}y + b_{2}c_{1} = 0

a_{2}b_{1}x + b_{1}b_{2}y + b_{1}c_{2} = 0

Subtracting, a_{1}b_{2}x - a_{2}b_{1}x + b_{2}c_{1} - b_{1}c_{2} = 0

or, x(a_{1}b_{2} - a_{2}b_{1}) = (b_{1}c_{2} - b_{2}c_{1})

or, x = (b_{1}c_{2} - b_{2}c_{1})/(a_{1}b_{2} - a_{2}b_{1})

Therefore, x/(b_{1}c_{2} - b_{2}c_{1}) = 1/(a_{1}b_{2} - a_{2}b_{1}) where (a_{1}b_{2} - a_{2}b_{1}) ≠ 0 -------------- (iv)

From equations (iii) and (iv), we get:

x/(b_{1}c_{2} - b_{2}c_{1}) = y/(c_{1}a_{2}) - c_{2}a_{1} = 1/(a_{1}b_{2} - a_{2}b_{1}) where (a_{1}b_{2} - a_{2}b_{1}) ≠ 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 (a_{1}b_{2} - a_{2}b_{1}) ≠ 0 from the two simultaneous linear equations

a_{1}x + b_{1}y + c_{1} = 0 ----------- (i)

a_{2}x + b_{2}y + c_{2} = 0 ----------- (ii)

we get, by the cross-multiplication method:

x/(b_{1}c_{2} - b_{2}c_{1}) = y/(c_{1}a_{2} - c_{2}a_{1}) = 1/(a_{1}b_{2} - a_{2}b_{1}) ---------- (A)

That means, x = (b_{1}c_{2} - b_{2}c_{1})/(a_{1}b_{2} - a_{2}b_{1})

y = (c_{1}a_{2} - c_{2}a_{1})/(a_{1}b_{2} - a_{2}b_{1})

**Note:**

If the value of x or y is zero, that is, (b_{1}c_{2} - b_{2}c_{1}) = 0 or (c_{1}a_{2} - c_{2}a_{1}) = 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 – c^{2} = 0

a^{2}x + b^{2}y – c^{2} = 0

**Solution:**

x/(-b + b^{2}) = y/(- a^{2} + a) = c^{2}/(ab^{2} - a^{2}b)

or, x/-b(1 - b) = y/- a(a - 1) = c^{2}/-ab(a - b)

or, x/b(1 - b) = y/a(a - 1) = c^{2}/ab(a - b)

or, x = bc^{2}(1 – b)/ab(a – b) = c^{2}(1 – b)/a(a – b) and y = c^{2}a(a – 1)/ab(a – b) = c^{2}(a – 1)/b(a – b)

Hence the required solution is:

x = c^{2}(1 – b)/a(a – b)

y = c^{2}a(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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