Cross Multiplication Method For Solving Equations

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

a₁x + b₁y + c₁ = 0                    ….(1)

a₂x + b₂y + c₂ = 0                             ….(2)

Multiplying equation (1) by b₂ and equation (2) by b₁, we get

a₁b₂x + b₁b₂y + b₂c₁ = 0              ….(3)

a₂b₁x + b₁b₂y + b₁c₂ = 0        ….(4)

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

(a₁b₂ – a₂b₁) x + (b₂c₁ – b₁c₂) = 0

(Rightarrow x=frac-}-})

(left[ -ne 0text{   and    }frac}}ne frac}} right])

(text{Similarly, }y=frac-}-})

These values of x and y can also be written as

(frac{x}-}=frac{-y}-}=frac{1}-})

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



(Rightarrow frac{x}{3times 14-5times 8}=frac{x}{3times 14-5times 8}=frac{1}{2times 5-4times 3})

(Rightarrow frac{x}{42-40}=frac{-y}{28-32}=frac{1}{10-12} )

(Rightarrow frac{x}{2}=frac{-y}{-4}=frac{1}{-2} )

(Rightarrow frac { x }{ 2 }) = (frac { -1 }{ 2 })

⇒ x = – 1

(Rightarrow frac { -y }{ -4 }) = (frac { -1 }{ 2 })

⇒ 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)

Using formula for cross multiplication method:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{x}{-6times 13-(-7)times 10}=frac{-y}{2times 13-3times 10}=frac{1}{2times (-7)-3times (-6)} )

(Rightarrow frac{x}{78+70}=frac{-y}{26-30}=frac{1}{-14+18} )

(Rightarrow frac{x}{-8}=frac{-y}{-4}=frac{1}{4} )

(Rightarrow frac { x }{ -8 }) = (frac { 1 }{ 4 })

⇒ x = – 2

(Rightarrow frac { -y }{ -4 }) = (frac { 1 }{ 4 })

⇒ 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

Using formula for cross multiplication method:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{x}{15times (-20)-(-2)times 23}=frac{-y}{11times (-20)-7times 23}=frac{1}{11times (-2)-7times 15})

(Rightarrow frac{x}{-300+46}=frac{-y}{-220-161}=frac{1}{-22-105} )

(Rightarrow frac{x}{-254}=frac{-y}{-381}=frac{1}{-127} )

(Rightarrow frac{x}{-254}=frac{1}{-127}Rightarrow x=2 )

(text{and}frac{-y}{-381}=frac{1}{-127}Rightarrow text{y}=text{ }-3 )

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

Using formula for cross multiplication method:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{x}{btimes {-(a+b)}-(-a)times {-(a-b)}}=frac{-y}{-a(a+b)+b(a-b)}=frac{1}{–} )

(Rightarrow frac{x}{-ab–+ab}=frac{-y}{–ab+ab-}=frac{1}{-(+)} )

(Rightarrow frac{x}{-(+)}=frac{-y}{-(+)}=frac{1}{-(+)} )

(Rightarrow frac{x}{-(+)}frac{1}{-(+)}Rightarrow x=1 )

(andtext{ }frac{-y}{-(+)}frac{1}{-(+)}Rightarrow y=-1 )

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

x + y = a – b;   ax – by = a² + b²

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

x + y – (a – b) = 0

ax – by – (a² + b²) = 0

Using formula for cross multiplication method:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{x}{-(+)-(-b)times {-(a-b)}}=frac{-y}{-(+)-atimes {-(a-b)}}=frac{1}{-b-a} )

(Rightarrow frac{x}{-(+)-b(a-b)}=frac{-y}{-(+)+a(a-b)}=frac{1}{-(b+a)} )

(Rightarrow frac{x}{—ab+}=frac{-y}{–+-ab}=frac{1}{-(a+b)} )

(Rightarrow frac{x}{-a(a+b)}=frac{-y}{-b(a+b)}=frac{1}{-(a+b)} )

(Rightarrow frac{x}{-a(a+b)}=frac{1}{-(a+b)}Rightarrow x=a )

(andtext{ }frac{-y}{-b(a+b)}=frac{1}{-(a+b)}Rightarrow y=-b )

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

(frac{x}{a}+frac{y}{b}=a+b ) ;   (frac{x}}+frac{y}}=2 )

Sol:    The given system of equations is rewritten as:

(frac{x}{a}+frac{y}{b}-left( a+b right))              ….(1)

(frac{x}}+frac{y}}-2 )             ….(2)

Multiplying equation (1) by ab, we get

bx + ay – ab (a + b) = 0              ….(3)

Multiplying equation (2) by a² b², we get

b²x + a²y – 2a²b² = 0                          ….(4)

Using formula for cross multiplication m
ethod:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{x}{-2+b(a+b)}=frac{-y}{-2+a(a+b)}=frac{1}b-a} )

(Rightarrow frac{x}{-2+b+}=frac{y}{-2++a}=frac{1}{ab(a-b)} )

(Rightarrow frac{x}b-}=frac{-y}{a-}=frac{1}{ab(a-b)} )

(Rightarrow frac{x}b(a-b)}=frac{y}{a(a-b)}=frac{1}{ab(a-b)} )

(Rightarrow frac{x}b(a-b)}=frac{1}{ab(a-b)} )

(Rightarrow x=fracb(a-b)}{ab(a-b)}= )

(Andtext{ }frac{y}{a(a-b)}=frac{1}{ab(a-b)} )

(Rightarrow y=frac{a(a-b)}{ab(a-b)}= )

Hence, the solution x = a², y = b²

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

ax + by = 1;   bx + ay = (frac}+}-1)

Sol:    The given system of equations can be written as

ax + by – 1 = 0            ….(1)

(bx+ay=frac}+}-1 )

(Rightarrow bx+ay=frac+2ab+–}+} )

(Rightarrow bx+ay=frac{2ab}+} )

(Rightarrow bx+ay-frac{2ab}+}=0 )        ….. (2)

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

ax + by – 1 = 0

(Rightarrow bx+ay-frac{2ab}+}=0 )

Using formula for cross multiplication method:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{x}{btimes left( frac{-2ab}+} right)-atimes (-1)}=frac{-y}{atimes left( frac{-2ab}+} right)-btimes (-1)}=frac{1}{atimes a-btimes b} )

(Rightarrow frac{x}{-frac{2a}+}+a}=frac{-y}{frac{-2b}+}+b}=frac{1}-} )

(Rightarrow frac{x}{frac{-2a++a}+}}=frac{-y}{frac{-2b+b+}+}}=frac{1}-} )

(Rightarrow frac{x}{frac{a(-)}+}}=frac{-y}{frac{b(-)}+}}=frac{1}-} )

( Rightarrow frac{x}{frac{a(-)}+}}=frac{1}-}Rightarrow x=frac{a}+} )

(andtext{ }frac{-y}{frac{b(-)}+}}=frac{1}-}Rightarrow y=frac{b}+} )

Hence, the solution is   (x=frac{a}+},y=frac{b}+} )

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

(a – b) x + (a + b) y = a² – 2ab – b²

(a + b) (x + y) = a² + b²

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

(a – b) x + (a +b) y – (a² – 2ab – b²) = 0

(a + b) x + (a + b) y – (a² + b²) = 0

Using formula for cross multiplication method:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{x}{(a+b)times {-(+)}-(a+b)times {-(-2ab-)}}=frac{-y}{(a-b)times {-(+)}-(a+b)times {-(-2ab-)}}=frac{1}{(a-b)times (a+b)-(a+b)times (a+b)} )

(Rightarrow frac{x}{-(a+b)(+)+(a+b)(-2ab-)}=frac{-y}{-(a-b)(+)+(a+b)(-2ab-)}=frac{1}{(a-b)(a+b)-} )

(Rightarrow frac{x}{(a+b)[-(+)+(a+b)(-2ab-)]}=frac{-y}{(a+b)(-2ab-)-(a-b)(+)}=frac{1}{(a+b)(a-b-a-b)} )

(Rightarrow frac{x}{(a+b)(-2ab-2)}=frac{-y}-b-3a—a+b+}=frac{1}{(a+b)(-2b)} )

(Rightarrow frac{x}{-(a+b)(2a+2b)b}=frac{-y}{-4a}=frac{1}{-2b(a+b)} )

(Rightarrow frac{x}{-2(a+b)(a+b)b}=frac{1}{-2b(a+b)}Rightarrow x=a+b )

(andtext{ }frac{-y}{-4a}=frac{1}{-2b(a+b)}Rightarrow y=frac{2ab}{a+b} )

Hence, the solution of the given system of equations is x = a + b, (y=frac{2ab}{a+b} )

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

a(x + y) + b (x – y) = a² – ab + b²

a(x + y) – b (x – y) = a² + ab + b²

Sol:    The given system of equations can be rewritten as

ax + bx + ay – by – ( a² – ab + b²) = 0

⇒ (a + b) x + (a – b) y – (a² – ab + b²) = 0 ….(1)

And  ax – bx + ay + by – (a² + ab + b²) = 0

⇒ (a – b) x + (a + b) y – (a² + ab + b²) = 0 …(2)

Using formula for cross multiplication method:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{x}{(a-b)times {-(+ab+)}-(a+b)times {-(-ab+)}}=frac{-y}{(a+b)times {-(+ab+)}-(a-b)times {-(-ab+)}}=frac{1}{(a+b)times (a+b)-(a-b)(a-b)} )

(Rightarrow frac{x}{-(a-b)(+ab+)+(a+b)(-ab+)}=frac{-y}{-(a+b)(+ab+)+(a-b)(-ab+)}=frac{1}-})

(Rightarrow frac{x}{-(-)+(+)}=frac{-y}{–2b-2a-+-2b+2a-}=frac{1}+2ab+-+2ab-})

(Rightarrow frac{x}{2}=frac{-y}{-4b-2}=frac{1}{4ab} )

(Rightarrow frac{x}{2}=frac{-y}{-2b(2+)}=frac{1}{4ab})

(Rightarrow frac{x}{2}=frac{1}{4ab}Rightarrow x=frac}{2a})

(andtext{ }frac{-y}{-2b(2+)}=frac{1}{4ab}Rightarrow y=frac{2+}{2a})

Hence, the solution is (x=frac}{2a},y=frac{2+}{2a})

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

(frac{a}{x}-frac{b}{y}=0;text{ }frac{a}{x}+fracb}{y}=+;)

Where x ≠ 0, y ≠ 0

Sol:    The given system of equations is

(frac{a}{x}-frac{b}{y}=0)     ………(1)

(frac{a}{x}+fracb}{y}-left( + right)=0) ………(2)

Putting  (frac { a }{ x }=u) and (frac { b }{ y }=v)  in equatinos (1) and (2) the system of equations reduces to

u – v + 0 = 0

b²u + a²v – (a² + b²) = 0

Using formula for cross multiplication method:



So, from equation (1) and (2) we can write the value of a,b and c.

(Rightarrow frac{u}+-times 0}=frac{-v}{-(+)-times 0}=frac{1}-(-)})

(Rightarrow frac{u}+}=frac{-v}{-(+)}=frac{1}+} )

(Rightarrow frac{u}+}=frac{1}+}Rightarrow u=1 )

(andtext{ }frac{-v}{-(+)}=frac{1}+}Rightarrow v=1 andtext{ u}=frac{a}{x}=1Rightarrow x=a )

(v=frac{b}{y}=1Rightarrow y=b )

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