20 solve for x and y using cross multiplication method x+2y 3 3x − 2y 7 0

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

\(\Rightarrow x=\frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\) \(\left[ {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}\ne 0\text{ and }\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}} \right]\) \(\text{Similarly, }y=\frac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\) These values of x and y can also be written as

\(\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{-y}{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\)

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

\(\Rightarrow \frac{x}{3\times 14-5\times 8}=\frac{x}{3\times 14-5\times 8}=\frac{1}{2\times 5-4\times 3}\) \(\Rightarrow \frac{x}{42-40}=\frac{-y}{28-32}=\frac{1}{10-12} \) \(\Rightarrow \frac{x}{2}=\frac{-y}{-4}=\frac{1}{-2} \) \(\Rightarrow \frac { x }{ 2 }\) = \(\frac { -1 }{ 2 }\) ⇒ 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. \(\Rightarrow \frac{x}{-6\times 13-(-7)\times 10}=\frac{-y}{2\times 13-3\times 10}=\frac{1}{2\times (-7)-3\times (-6)} \) \(\Rightarrow \frac{x}{78+70}=\frac{-y}{26-30}=\frac{1}{-14+18} \) \(\Rightarrow \frac{x}{-8}=\frac{-y}{-4}=\frac{1}{4} \) \(\Rightarrow \frac { x }{ -8 }\) = \(\frac { 1 }{ 4 }\) ⇒ 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. \(\Rightarrow \frac{x}{15\times (-20)-(-2)\times 23}=\frac{-y}{11\times (-20)-7\times 23}=\frac{1}{11\times (-2)-7\times 15}\) \(\Rightarrow \frac{x}{-300+46}=\frac{-y}{-220-161}=\frac{1}{-22-105} \) \(\Rightarrow \frac{x}{-254}=\frac{-y}{-381}=\frac{1}{-127} \) \(\Rightarrow \frac{x}{-254}=\frac{1}{-127}\Rightarrow x=2 \) \(\text{and}\frac{-y}{-381}=\frac{1}{-127}\Rightarrow \text{y}=\text{ }-3 \)

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

\(and\text{ }\frac{-y}{-({{a}^{2}}+{{b}^{2}})}\frac{1}{-({{a}^{2}}+{{b}^{2}})}\Rightarrow y=-1 \)

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

\(and\text{ }\frac{-y}{-b(a+b)}=\frac{1}{-(a+b)}\Rightarrow y=-b \)

Example 6: Solve the following system of equations by the method of cross-multiplication: \(\frac{x}{a}+\frac{y}{b}=a+b \) ; \(\frac{x}{{{a}^{2}}}+\frac{y}{{{b}^{2}}}=2 \)

Sol: The given system of equations is rewritten as:

\(\frac{x}{a}+\frac{y}{b}-\left( a+b \right)\) ….(1) \(\frac{x}{{{a}^{2}}}+\frac{y}{{{b}^{2}}}-2 \) ….(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 method:

So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{-2{{a}^{3}}{{b}^{2}}+{{a}^{3}}b(a+b)}=\frac{-y}{-2{{a}^{2}}{{b}^{3}}+a{{b}^{3}}(a+b)}=\frac{1}{{{a}^{2}}b-a{{b}^{2}}} \) \(\Rightarrow \frac{x}{-2{{a}^{3}}{{b}^{2}}+{{a}^{4}}b+{{a}^{3}}{{b}^{2}}}=\frac{y}{-2{{a}^{2}}{{b}^{3}}+{{a}^{2}}{{b}^{3}}+a{{b}^{4}}}=\frac{1}{ab(a-b)} \) \(\Rightarrow \frac{x}{{{a}^{4}}b-{{a}^{3}}{{b}^{2}}}=\frac{-y}{a{{b}^{4}}-{{a}^{2}}{{b}^{3}}}=\frac{1}{ab(a-b)} \) \(\Rightarrow \frac{x}{{{a}^{3}}b(a-b)}=\frac{y}{a{{b}^{3}}(a-b)}=\frac{1}{ab(a-b)} \) \(\Rightarrow \frac{x}{{{a}^{3}}b(a-b)}=\frac{1}{ab(a-b)} \) \(\Rightarrow x=\frac{{{a}^{3}}b(a-b)}{ab(a-b)}={{a}^{2}} \) \(And\text{ }\frac{y}{a{{b}^{3}}(a-b)}=\frac{1}{ab(a-b)} \) \(\Rightarrow y=\frac{a{{b}^{3}}(a-b)}{ab(a-b)}={{b}^{2}} \)

Hence, the solution x = a2, y = b2

Example 7: Solve the following system of equations by cross-multiplication method – ax + by = 1; bx + ay = \(\frac{{{(a+b)}^{2}}}{{{a}^{2}}+{{b}^{2}}}-1\)

Sol: The given system of equations can be written as

ax + by – 1 = 0 ….(1) \(bx+ay=\frac{{{(a+b)}^{2}}}{{{a}^{2}}+{{b}^{2}}}-1 \) \(\Rightarrow bx+ay=\frac{{{a}^{2}}+2ab+{{b}^{2}}-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \) \(\Rightarrow bx+ay=\frac{2ab}{{{a}^{2}}+{{b}^{2}}} \) \(\Rightarrow bx+ay-\frac{2ab}{{{a}^{2}}+{{b}^{2}}}=0 \) ….. (2) Rewritting the equations (1) and (2), we have ax + by – 1 = 0 \(\Rightarrow bx+ay-\frac{2ab}{{{a}^{2}}+{{b}^{2}}}=0 \) Using formula for cross multiplication method:

So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{b\times \left( \frac{-2ab}{{{a}^{2}}+{{b}^{2}}} \right)-a\times (-1)}=\frac{-y}{a\times \left( \frac{-2ab}{{{a}^{2}}+{{b}^{2}}} \right)-b\times (-1)}=\frac{1}{a\times a-b\times b} \) \(\Rightarrow \frac{x}{-\frac{2a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}+a}=\frac{-y}{\frac{-2{{a}^{2}}b}{{{a}^{2}}+{{b}^{2}}}+b}=\frac{1}{{{a}^{2}}-{{b}^{2}}} \) \(\Rightarrow \frac{x}{\frac{-2a{{b}^{2}}+{{a}^{3}}+a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}=\frac{-y}{\frac{-2{{a}^{2}}b+{{a}^{2}}b+{{b}^{3}}}{{{a}^{2}}+{{b}^{2}}}}=\frac{1}{{{a}^{2}}-{{b}^{2}}} \) \(\Rightarrow \frac{x}{\frac{a({{a}^{2}}-{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}}=\frac{-y}{\frac{b({{b}^{2}}-{{a}^{2}})}{{{a}^{2}}+{{b}^{2}}}}=\frac{1}{{{a}^{2}}-{{b}^{2}}} \) \( \Rightarrow \frac{x}{\frac{a({{a}^{2}}-{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}}=\frac{1}{{{a}^{2}}-{{b}^{2}}}\Rightarrow x=\frac{a}{{{a}^{2}}+{{b}^{2}}} \) \(and\text{ }\frac{-y}{\frac{b({{b}^{2}}-{{a}^{2}})}{{{a}^{2}}+{{b}^{2}}}}=\frac{1}{{{a}^{2}}-{{b}^{2}}}\Rightarrow y=\frac{b}{{{a}^{2}}+{{b}^{2}}} \)

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

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. \(\Rightarrow \frac{x}{(a+b)\times \{-({{a}^{2}}+{{b}^{2}})\}-(a+b)\times \{-({{a}^{2}}-2ab-{{b}^{2}})\}}=\frac{-y}{(a-b)\times \{-({{a}^{2}}+{{b}^{2}})\}-(a+b)\times \{-({{a}^{2}}-2ab-{{b}^{2}})\}}=\frac{1}{(a-b)\times (a+b)-(a+b)\times (a+b)} \) \(\Rightarrow \frac{x}{-(a+b)({{a}^{2}}+{{b}^{2}})+(a+b)({{a}^{2}}-2ab-{{b}^{2}})}=\frac{-y}{-(a-b)({{a}^{2}}+{{b}^{2}})+(a+b)({{a}^{2}}-2ab-{{b}^{2}})}=\frac{1}{(a-b)(a+b)-{{(a+b)}^{2}}} \) \(\Rightarrow \frac{x}{(a+b)[-({{a}^{2}}+{{b}^{2}})+(a+b)({{a}^{2}}-2ab-{{b}^{2}})]}=\frac{-y}{(a+b)({{a}^{2}}-2ab-{{b}^{2}})-(a-b)({{a}^{2}}+{{b}^{2}})}=\frac{1}{(a+b)(a-b-a-b)} \) \(\Rightarrow \frac{x}{(a+b)(-2ab-2{{b}^{2}})}=\frac{-y}{{{a}^{3}}-{{a}^{2}}b-3a{{b}^{2}}-{{b}^{3}}-{{a}^{3}}-a{{b}^{2}}+{{a}^{2}}b+{{b}^{3}}}=\frac{1}{(a+b)(-2b)} \) \(\Rightarrow \frac{x}{-(a+b)(2a+2b)b}=\frac{-y}{-4a{{b}^{2}}}=\frac{1}{-2b(a+b)} \) \(\Rightarrow \frac{x}{-2(a+b)(a+b)b}=\frac{1}{-2b(a+b)}\Rightarrow x=a+b \) \(and\text{ }\frac{-y}{-4a{{b}^{2}}}=\frac{1}{-2b(a+b)}\Rightarrow y=\frac{2ab}{a+b} \)

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. \(\Rightarrow \frac{x}{(a-b)\times \{-({{a}^{2}}+ab+{{b}^{2}})\}-(a+b)\times \{-({{a}^{2}}-ab+{{b}^{2}})\}}=\frac{-y}{(a+b)\times \{-({{a}^{2}}+ab+{{b}^{2}})\}-(a-b)\times \{-({{a}^{2}}-ab+{{b}^{2}})\}}=\frac{1}{(a+b)\times (a+b)-(a-b)(a-b)} \) \(\Rightarrow \frac{x}{-(a-b)({{a}^{2}}+ab+{{b}^{2}})+(a+b)({{a}^{2}}-ab+{{b}^{2}})}=\frac{-y}{-(a+b)({{a}^{2}}+ab+{{b}^{2}})+(a-b)({{a}^{2}}-ab+{{b}^{2}})}=\frac{1}{{{(a+b)}^{2}}-{{(a-b)}^{2}}}\) \(\Rightarrow \frac{x}{-({{a}^{3}}-{{b}^{3}})+({{a}^{3}}+{{b}^{2}})}=\frac{-y}{-{{a}^{3}}-2{{a}^{2}}b-2a{{b}^{2}}-{{b}^{3}}+{{a}^{3}}-2{{a}^{2}}b+2a{{b}^{2}}-{{b}^{3}}}=\frac{1}{{{a}^{2}}+2ab+{{b}^{2}}-{{a}^{2}}+2ab-{{b}^{2}}}\) \(\Rightarrow \frac{x}{2{{b}^{3}}}=\frac{-y}{-4{{a}^{2}}b-2{{b}^{3}}}=\frac{1}{4ab} \) \(\Rightarrow \frac{x}{2{{b}^{3}}}=\frac{-y}{-2b(2{{a}^{2}}+{{b}^{2}})}=\frac{1}{4ab}\) \(\Rightarrow \frac{x}{2{{b}^{3}}}=\frac{1}{4ab}\Rightarrow x=\frac{{{b}^{2}}}{2a}\) \(and\text{ }\frac{-y}{-2b(2{{a}^{2}}+{{b}^{2}})}=\frac{1}{4ab}\Rightarrow y=\frac{2{{a}^{2}}+{{b}^{2}}}{2a}\)

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

Example 10: Solve the following system of equations by the method of cross-multiplication. \(\frac{a}{x}-\frac{b}{y}=0;\text{ }\frac{a{{b}^{2}}}{x}+\frac{{{a}^{2}}b}{y}={{a}^{2}}+{{b}^{2}};\) Where x ≠ 0, y ≠ 0

Sol: The given system of equations is

\(\frac{a}{x}-\frac{b}{y}=0\) ………(1) \(\frac{a{{b}^{2}}}{x}+\frac{{{a}^{2}}b}{y}-\left( {{a}^{2}}+{{b}^{2}} \right)=0\) ………(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. \(\Rightarrow \frac{u}{{{a}^{2}}+{{b}^{2}}-{{a}^{2}}\times 0}=\frac{-v}{-({{a}^{2}}+{{b}^{2}})-{{b}^{2}}\times 0}=\frac{1}{{{a}^{2}}-(-{{b}^{2}})}\) \(\Rightarrow \frac{u}{{{a}^{2}}+{{b}^{2}}}=\frac{-v}{-({{a}^{2}}+{{b}^{2}})}=\frac{1}{{{a}^{2}}+{{b}^{2}}} \) \(\Rightarrow \frac{u}{{{a}^{2}}+{{b}^{2}}}=\frac{1}{{{a}^{2}}+{{b}^{2}}}\Rightarrow u=1 \) \(and\text{ }\frac{-v}{-({{a}^{2}}+{{b}^{2}})}=\frac{1}{{{a}^{2}}+{{b}^{2}}}\Rightarrow v=1 and\text{ u}=\frac{a}{x}=1\Rightarrow x=a \) \(v=\frac{b}{y}=1\Rightarrow y=b \)

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