Question

Solve by cross-multiplication method:
$\left( {a - b} \right)x + \left( {a + b} \right)y = 2\left( {{a^2} - {b^2}} \right)$
$\left( {a + b} \right)x - \left( {a - b} \right)y = 4ab$
A) $x = \left( {a - b} \right)$ and $y = \left( {2a - b} \right)$
B) $x = \left( {a + b} \right)$ and $y = \left( {a - b} \right)$
C) $x = \left( {2a + 5b} \right)$ and $y = \left( {a - {b^2}} \right)$
D) $x = \left( {2a - 3b} \right)$ and $y = \left( {3a - b} \right)$

Answer 1

Hint: To solve the problem first we have to take all the terms to the left hand side of the equations and convert them in the form,
${a_1}x + {b_1}y + {c_1} = 0$
${a_2}x + {b_2}y + {c_2} = 0$
Then, we are to apply the formula of cross multiplication, that is,
$\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
Then, equating the $x$ term and the constant term we will get the value of $x$. Similarly, equating the $y$ term and the constant term we will get the value of $y$.

Complete step by step solution:
The two given equations are,
$\left( {a - b} \right)x + \left( {a + b} \right)y = 2\left( {{a^2} - {b^2}} \right)$
$\left( {a + b} \right)x - \left( {a - b} \right)y = 4ab$
We can rewrite the equations in the form suitable for cross-multiplication as,
$\left( {a - b} \right)x + \left( {a + b} \right)y - 2\left( {{a^2} - {b^2}} \right) = 0 - - - \left( 1 \right)$
$\left( {a + b} \right)x - \left( {a - b} \right)y - 4ab = 0 - - - \left( 2 \right)$
[Taking all the terms to the left hand side]
The general form of equations for cross-multiplication is,
${a_1}x + {b_1}y + {c_1} = 0$
${a_2}x + {b_2}y + {c_2} = 0$
Comparing the equations $\left( 1 \right)$ and $\left( 2 \right)$ to the general equations, we get,
${a_1} = \left( {a - b} \right),{b_1} = \left( {a + b} \right),{c_1} = - 2\left( {{a^2} - {b^2}} \right)$
${a_2} = \left( {a + b} \right),{b_2} = - \left( {a - b} \right),{c_2} = - 4ab$
The formula of cross-multiplication is,
$\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
Substituting the values in the formula, we get,
$\dfrac{x}{{\left( {a + b} \right)\left( { - 4ab} \right) - \left( { - \left( {a - b} \right)} \right)\left( { - 2\left( {{a^2} - {b^2}} \right)} \right)}} = \dfrac{y}{{\left( { - 2\left( {{a^2} - {b^2}} \right)} \right)\left( {a + b} \right) - \left( { - 4ab} \right)\left( {a - b} \right)}} = \dfrac{1}{{\left( {a - b} \right)\left( { - \left( {a - b} \right)} \right) - \left( {a + b} \right)\left( {a + b} \right)}}$$ \Rightarrow \dfrac{x}{{ - 4ab\left( {a + b} \right) - 2\left( {a + b} \right){{\left( {a - b} \right)}^2}}} = \dfrac{y}{{4ab\left( {a - b} \right) - 2\left( {a + b} \right){{\left( {a - b} \right)}^2}}} = \dfrac{1}{{ - {{\left( {a - b} \right)}^2} - {{\left( {a + b} \right)}^2}}}$
[Using, $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$]
Taking terms common in the denominator, we get,
$ \Rightarrow \dfrac{x}{{ - \left( {a + b} \right)\left( {4ab + 2{{\left( {a - b} \right)}^2}} \right)}} = \dfrac{y}{{\left( {a - b} \right)\left( {4ab - 2\left( {a + b} \right)\left( {a + b} \right)} \right)}} = \dfrac{1}{{ - \left( {{{\left( {a - b} \right)}^2} + {{\left( {a + b} \right)}^2}} \right)}}$
Now, simplifying, we get,
$ \Rightarrow \dfrac{x}{{ - \left( {a + b} \right)\left( {4ab + 2\left( {{a^2} - 2ab + {b^2}} \right)} \right)}} = \dfrac{y}{{\left( {a - b} \right)\left( {4ab - 2{{\left( {a + b} \right)}^2}} \right)}} = \dfrac{1}{{ - \left( {\left( {{a^2} - 2ab + {b^2}} \right) + \left( {{a^2} + {b^2} + 2ab} \right)} \right)}}$
[Using, \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]]
Now, opening the brackets and simplifying, we get,
$ \Rightarrow \dfrac{x}{{ - \left( {a + b} \right)\left( {4ab + 2{a^2} - 4ab + 2{b^2}} \right)}} = \dfrac{y}{{\left( {a - b} \right)\left( {4ab - 2{a^2} - 2{b^2} - 4ab} \right)}} = \dfrac{1}{{ - \left( {2\left( {{a^2} + {b^2}} \right)} \right)}}$
[Using, \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]]
$ \Rightarrow \dfrac{x}{{ - \left( {a + b} \right)\left( {2{a^2} + 2{b^2}} \right)}} = \dfrac{y}{{\left( {a - b} \right)\left( { - 2{a^2} - 2{b^2}} \right)}} = \dfrac{1}{{ - \left( {2\left( {{a^2} + {b^2}} \right)} \right)}}$
Now, equating the $x$ term and the constant term, we get,
$\dfrac{x}{{ - \left( {a + b} \right)\left( {2{a^2} + 2{b^2}} \right)}} = \dfrac{1}{{ - \left( {2\left( {{a^2} + {b^2}} \right)} \right)}}$
Taking $2$ common in the denominator of the $x$ term, we get,
$ \Rightarrow \dfrac{x}{{ - 2\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)}} = \dfrac{1}{{ - \left( {2\left( {{a^2} + {b^2}} \right)} \right)}}$
Cancelling the similar terms, we get,
$ \Rightarrow \dfrac{x}{{\left( {a + b} \right)}} = \dfrac{1}{1}$
Multiplying both sides with $\left( {a + b} \right)$, we get,
$ \Rightarrow x = \left( {a + b} \right)$
Now, equating the $y$ term and the constant term, we get,
$\dfrac{y}{{\left( {a - b} \right)\left( { - 2{a^2} - 2{b^2}} \right)}} = \dfrac{1}{{ - \left( {2\left( {{a^2} + {b^2}} \right)} \right)}}$
Taking $ - 2$ common in the denominator of the $y$ term, we get,
$ \Rightarrow \dfrac{y}{{ - 2\left( {a - b} \right)\left( {{a^2} + {b^2}} \right)}} = \dfrac{1}{{ - \left( {2\left( {{a^2} + {b^2}} \right)} \right)}}$
Cancelling the similar terms, we get,
$ \Rightarrow \dfrac{y}{{\left( {a - b} \right)}} = \dfrac{1}{1}$
Multiplying both sides with $\left( {a - b} \right)$, we get,
$ \Rightarrow y = \left( {a - b} \right)$
Therefore, the required values are,
$x = \left( {a + b} \right),y = \left( {a - b} \right)$
The correct option is B.
So, the correct answer is “Option B”.

Note: -Cross multiplication is another method of solving two linear equations in two variables other than elimination method and substitution method. The same question can also be solved by elimination or substitution method, but as mentioned in the question, we have to solve by cross-multiplication method.