**Cross Multiplication Method For Solving Equations**

By the method of elimination by substitution, only those equations can be solved, which have unique solution. But the method of cross multiplication discussed below is applicable in all the cases; whether the system has a unique solution, no solution or infinitely many solutions.

Let us solve the following system of equations

a_{1}x + b_{1}y + c_{1} = 0 ….(1)

a_{2}x + b_{2}y + c_{2} = 0 ….(2)

Multiplying equation (1) by b_{2} and equation (2) by b_{1}, we get

a_{1}b_{2}x + b_{1}b_{2}y + b_{2}c_{1} = 0 ….(3)

a_{2}b_{1}x + b_{1}b_{2}y + b_{1}c_{2} = 0 ….(4)

Subtracting equation (4) from equation (3), we get

(a_{1}b_{2} – a_{2}b_{1}) x + (b_{2}c_{1} – b_{1}c_{2}) = 0

These values of x and y can also be written as

**Cross Multiplication Method Examples**

**Example 1: ** Solve the following system of equations by cross-multiplication method.

2x + 3y + 8 = 0

4x + 5y + 14 = 0

**Sol. ** The given system of equations is

2x + 3y + 8 = 0

4x + 5y + 14 = 0

By cross-multiplication, we get

=

⇒ x = – 1

=

⇒ y = – 2

Hence, the solution is x = – 1, y = – 2

We can verify the solution.

**Example 2: ** Solve the follownig system of equations by the method of cross-multiplication.

2x – 6y + 10 = 0

3x – 7y + 13 = 0

**Sol.** The given system of equations is

2x – 6y + 10 = 0 ….(1)

3x – 7y + 13 = 0 ….(2)

By cross-multiplication, we have

=

⇒ x = – 2

=

⇒ y = 1

Hence, the solution is x = – 2, y = 1

**Example 3: ** Solve the following system of equations by the method of cross-multiplication.

11x + 15y = – 23; 7x – 2y = 20

**Sol. ** The given system of equations is

11x + 15y + 23 = 0

7x – 2y – 20 = 0

Now, by cross-multiplication method, we have

Hence, x = 2, y = – 3 is the required solution.

**Example 4: ** Solve the following system of equations by cross-multiplication method.

ax + by = a – b; bx – ay = a + b

**Sol. ** Rewriting the given system of equations, we get

ax + by – (a – b) = 0

bx – ay – (a + b) = 0

By cross-multiplication method, we have

**Example 5: **Solve the following system of equations by cross-multiplication method.

x + y = a – b; ax – by = a^{2} + b^{2}

**Sol. ** The given system of equations can be rewritten as:

x + y – (a – b) = 0

ax – by – (a^{2} + b^{2}) = 0

By cross-multiplication method, we have

**Example 6: **Solve the following system of equations by the method of cross-multiplication:

;

**Sol: **The given system of equations is rewritten as:

….(1)

….(2)

Multiplying equation (1) by ab, we get

bx + ay – ab (a + b) = 0 ….(3)

Multiplying equation (2) by a^{2 }b^{2}, we get

b^{2}x + a^{2}y – 2a^{2}b^{2} = 0 ….(4)

By cross multiplication method, we have

Hence, the solution x = a^{2}, y = b^{2}

**Example 7: **Solve the following system of equations by cross-multiplication method –

ax + by = 1; bx + ay =

**Sol: **The given system of equations can be written as

ax + by – 1 = 0 ….(1)

….. (2)

Rewritting the equations (1) and (2), we have

ax + by – 1 = 0

Now, by cross-multiplication method, we have

Hence, the solution is

**Example 8: **Solve the following system of equations in x and y by cross-multiplication method

(a – b) x + (a + b) y = a^{2} – 2ab – b^{2}

(a + b) (x + y) = a^{2} + b^{2}

**Sol: **The given system of equations can be rewritten as :

(a – b) x + (a +b) y – (a^{2} – 2ab – b^{2}) = 0

(a + b) x + (a + b) y – (a^{2} + b^{2}) = 0

By cross-multiplication method, we have

Hence, the solution of the given system of equations is x = a + b,

**Example 9: **Solve the following system of equations by cross-multiplications method.

a(x + y) + b (x – y) = a^{2} – ab + b^{2}

a(x + y) – b (x – y) = a^{2} + ab + b^{2}

**Sol: **The given system of equations can be rewritten as

ax + bx + ay – by – ( a^{2} – ab + b^{2}) = 0

⇒ (a + b) x + (a – b) y – (a^{2} – ab + b^{2}) = 0 ….(1)

And ax – bx + ay + by – (a^{2} + ab + b^{2}) = 0

⇒ (a – b) x + (a + b) y – (a^{2} + ab + b^{2}) = 0 …(2)

Now, by cross-multiplication method, we have

Hence, the solution is

**Example 10: **Solve the following system of equations by the method of cross-multiplication.

Where x ≠ 0, y ≠ 0

**Sol: **The given system of equations is

………(1)

………(2)

Putting and in equatinos (1) and (2) the system of equations reduces to

u – v + 0 = 0

b^{2}u + a^{2}v – (a^{2} + b^{2}) = 0

By the method of cross-multiplication, we have

Hence, the solution of the given system of equations is x = a, y = b.

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Nishith Vaishnav says

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