Question

Solve the following pair of equations by cross-multiplication method:
\[\dfrac{x}{a} + \dfrac{y}{b} = 2\]
\[ax + by = {a^2} + {b^2}\]

Answer 1

Hint: Here we have given two equations and we have to find the solution of it by using a cross-multiplication method. The method states that if \[{l_1}x + {m_1}y + {n_1} = 0\] and \[{l_2}x + {m_2}y + {n_2} = 0\] are two linear equations where \[{l_1},{\text{ }}{m_1},{\text{ }}{n_1}\] are coefficient of \[x,{\text{ }}y\] and constant of equation (1) respectively and similarly \[{l_2},{\text{ }}{m_2},{\text{ }}{n_2}\] are coefficient of \[x,{\text{ }}y\] and constant of equation (2) respectively. To solve the above set of linear equations, we have to find the value of \[x\] and \[y\] variables using the following formulae-
\[\dfrac{x}{{{m_1}{n_2} - {m_2}{n_1}}} = {\text{ }}\dfrac{y}{{{n_1}{l_2} - {n_2}{l_1}}} = {\text{ }}\dfrac{1}{{{l_1}{m_2} - {l_2}{m_1}}}\]
So, in order to solve the given question, we will compare the given pair of equations with the general form and put the values in the formula. And finally, we will solve the values of \[x\] and \[y\] and get the desired result.

Complete step by step solution:
The given two equations are-
\[\dfrac{x}{a} + \dfrac{y}{b} = 2\]
\[ax + by = {a^2} + {b^2}\]
First of all, convert the above linear equations in the general form of the linear equation by shifting all the terms on one side of equality.
Therefore, we get
\[\dfrac{x}{a} + \dfrac{y}{b} - 2 = 0{\text{ }} - - - \left( 1 \right)\]
\[ax + by - \left( {{a^2} + {b^2}} \right) = 0{\text{ }} - - - \left( 2 \right)\]
Now, on comparing equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] with the general form i.e.,
\[{l_1}x + {m_1}y + {n_1} = 0\] and
\[{l_2}x + {m_2}y + {n_2} = 0\]
we get the following values of coefficients as,
\[{l_1} = \dfrac{1}{a},{\text{ }}{m_1} = \dfrac{1}{b},{\text{ }}{n_1} = - 2\]
\[{l_2} = a,{\text{ }}{m_2} = b,{\text{ }}{n_2} = - \left( {{a^2} + {b^2}} \right)\]
Now, we know that formula for cross-multiplication method is- \[\dfrac{x}{{{m_1}{n_2} - {m_2}{n_1}}} = {\text{ }}\dfrac{y}{{{n_1}{l_2} - {n_2}{l_1}}} = {\text{ }}\dfrac{1}{{{l_1}{m_2} - {l_2}{m_1}}}\]
So, on substituting the values of the coefficients in the above formula, we get
\[\dfrac{x}{{ - \dfrac{1}{b}\left( {{a^2} + {b^2}} \right) - b\left( { - 2} \right)}}{\text{ }} = {\text{ }}\dfrac{y}{{\left( { - 2} \right)a - \left( { - \left( {{a^2} + {b^2}} \right)} \right)\dfrac{1}{a}}}{\text{ }} = {\text{ }}\dfrac{1}{{\dfrac{1}{a}b - a\dfrac{1}{b}}}\]
Further simplifying it, we get
\[ \Rightarrow \dfrac{x}{{ - \dfrac{{\left( {{a^2} + {b^2}} \right)}}{b} + 2b}}{\text{ }} = {\text{ }}\dfrac{y}{{\left( { - 2} \right)a + \dfrac{{\left( {{a^2} + {b^2}} \right)}}{a}}}{\text{ }} = {\text{ }}\dfrac{1}{{\dfrac{b}{a} - \dfrac{a}{b}}}\]
On taking L.C.M, we get
\[ \Rightarrow \dfrac{x}{{\dfrac{{ - \left( {{a^2} + {b^2}} \right) + 2{b^2}}}{b}}}{\text{ }} = {\text{ }}\dfrac{y}{{\dfrac{{ - 2{a^2} + \left( {{a^2} + {b^2}} \right)}}{a}}}{\text{ }} = {\text{ }}\dfrac{1}{{\dfrac{{{b^2} - {a^2}}}{{ab}}}}\]
\[ \Rightarrow \dfrac{{bx}}{{ - \left( {{a^2} + {b^2}} \right) + 2{b^2}}}{\text{ }} = {\text{ }}\dfrac{{ay}}{{ - 2{a^2} + \left( {{a^2} + {b^2}} \right)}}{\text{ }} = {\text{ }}\dfrac{{ab}}{{{b^2} - {a^2}}}\]
On simplifying the denominators, we get
\[ \Rightarrow \dfrac{{bx}}{{{b^2} - {a^2}}}{\text{ }} = {\text{ }}\dfrac{{ay}}{{{b^2} - {a^2}}}{\text{ }} = {\text{ }}\dfrac{{ab}}{{{b^2} - {a^2}}}\]
Since, the denominator is not zero, so a unique solution exists.
Now, separating the above into two equalities, we get
\[\dfrac{{bx}}{{{b^2} - {a^2}}}{\text{ }} = {\text{ }}\dfrac{{ab}}{{{b^2} - {a^2}}}\] and \[\dfrac{{ay}}{{{b^2} - {a^2}}}{\text{ }} = {\text{ }}\dfrac{{ab}}{{{b^2} - {a^2}}}\]
On cancelling the denominators, we get
\[bx = ab\] and \[ay = ab\]
On solving for \[x\] and \[y\] we get
\[x = \dfrac{{ab}}{b}\] and \[y = \dfrac{{ab}}{a}\]
\[ \Rightarrow x = a\] and \[ \Rightarrow y = b\]
So, the solution of \[x\] and \[y\] are \[x = a\] and \[y = b\]
Hence, on solving the two pair of equations i.e., \[\dfrac{x}{a} + \dfrac{y}{b} = 2\] and \[ax + by = {a^2} + {b^2}\] we get the values as \[x = a\] and \[y = b\]
So, the correct answer is “\[x = a\] and \[y = b\]”.

Note: Cross multiplication method can only be applied when we have a pair of linear equations in two variables. In the cross multiplication method the representation is very crucial. Generally, students get confused while writing the formula which leads to the wrong answer. Be careful while comparing, you must make the right-hand side zero.
Also remember that there is another form of writing this formula which is,
\[\dfrac{x}{{{m_1}{n_2} - {m_2}{n_1}}} = {\text{ }}\dfrac{{ - y}}{{{l_1}{n_2} - {l_2}{n_1}}} = {\text{ }}\dfrac{1}{{{l_1}{m_2} - {l_2}{m_1}}}\]
Don’t get confused between the two, both are the same. To make it easier to remember we just changed the sign of \[y\] component in the formula we used to solve the question.