Question

Solve the following system of equations in $x$ and $y$ ,
$ax + by = 1$ and $bx + ay = \dfrac{{\mathop {(a + b)}\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }} - 1$ or $bx + ay = \dfrac{{2ab}}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$ .

Answer 1

Hint: We will be solving the above system of equations by the elimination method that is used to solve the system of linear equations with two variables. In the elimination method, we first write the equations in a standard format and then try to make the coefficient of any one of the two variables equal, and then we subtract the equations to eliminate one variable and then we solve the resultant to find the value for the other variable. After getting the value of the other variable, we will substitute that value in any one of the two equations and then we can get the value of the other variable as well.

Complete step by step solution:
We are given a system of two equations in $x$ and $y$ in the question that needs to be solved.
The equations are:
$ax + by = 1....................(i)$, and
$bx + ay = \dfrac{{\mathop {(a + b)}\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }} - 1$ or $bx + ay = \dfrac{{2ab}}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}....................(ii)$ .
To solve the system of equations using the elimination method we will first multiply $(i)$ by $\dfrac{1}{a}$ and $(ii)$ by $\dfrac{1}{b}$ .
On multiplying an equation $(i)$ by $\dfrac{1}{a}$ we get the equation as
$\dfrac{1}{a} \times (ax + by) = \dfrac{1}{a} \times 1$
Now let us simplify by multiplying the term outside the brackets on the left-hand side
$ \Rightarrow \dfrac{1}{a} \times ax + \dfrac{1}{a} \times by = \dfrac{1}{a}$
On further simplification we get
$ \Rightarrow x + \dfrac{b}{a}y = \dfrac{1}{a}..........................(iii)$
On multiplying the equation $(ii)$ by $\dfrac{1}{b}$ we get the equation as
$\dfrac{1}{b} \times (bx + ay) = \dfrac{1}{b} \times \left( {\dfrac{{2ab}}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}} \right)$
Now let us simplify by multiplying the term outside the brackets.
$ \Rightarrow \dfrac{1}{b} \times bx + \dfrac{1}{b} \times ay = \dfrac{{2a}}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$
On further simplification we get
$ \Rightarrow x + \dfrac{a}{b}y = \dfrac{{2a}}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}..........................(iv)$
Thus, we have made the coefficient of variable $x$ the same in both the equations so now we will subtract one equation from the other to eliminate the variable $x$ .
$\therefore $ Subtracting equation $(iv)$ from the equation $(iii)$ we get:
$
{\text{ }}x + \dfrac{b}{a}y = \dfrac{1}{a} \\
- \left( {x + \dfrac{a}{b}y = \dfrac{{2a}}{{(\mathop a\nolimits^2 + \mathop b\nolimits^2 )}}} \right) \\
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
{\text{ }}\dfrac{b}{a}y - \dfrac{a}{b}y = \dfrac{1}{a} - \dfrac{{2a}}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }} \\
$
Now we have to solve this equation to get the value of the variable $y$ .
On simplifying the equation $\left( {\dfrac{b}{a} - \dfrac{a}{b}} \right)y = \dfrac{1}{a} - \dfrac{{2a}}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$ we get
On taking LCM on both sides of the equation we get
\[ \Rightarrow \left( {\dfrac{{b \times b - a \times a}}{{ab}}} \right)y = \left( {\dfrac{{\mathop a\nolimits^2 + \mathop b\nolimits^2 - 2a \times a}}{{a \times \left( {\mathop a\nolimits^2 + \mathop b\nolimits^2 } \right)}}} \right)\]
Simplifying the above we get
\[ \Rightarrow \left( {\dfrac{{\mathop b\nolimits^2 - \mathop a\nolimits^2 }}{b}} \right)y = \left( {\dfrac{{\mathop a\nolimits^2 + \mathop b\nolimits^2 - 2\mathop a\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}} \right)\]
\[ \Rightarrow \left( {\dfrac{{\mathop b\nolimits^2 - \mathop a\nolimits^2 }}{b}} \right)y = \left( {\dfrac{{\mathop b\nolimits^2 - \mathop a\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}} \right)\]
\[ \Rightarrow \dfrac{y}{b} = \dfrac{1}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}\]
\[ \Rightarrow y = \dfrac{b}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}\]
Thus, we got the value of one variable to find the other we have to substitute it in anyone of the equation.
Let us substitute the value of $y = \dfrac{b}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$ in equation $(i)$ to solve for the variable $x$ .
Therefore, we get:
$ax + by = 1$ $ \Rightarrow ax + b \times \left( {\dfrac{b}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}} \right) = 1$
On simplifying the above, we get
$ \Rightarrow ax + \dfrac{{\mathop b\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }} = 1$
$ \Rightarrow ax = 1 - \dfrac{{\mathop b\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$
Now taking LCM on the right-hand side we get
$ \Rightarrow ax = \dfrac{{\mathop a\nolimits^2 + \mathop b\nolimits^2 - \mathop b\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$
On further simplification we get
$ \Rightarrow ax = \dfrac{{\mathop a\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$
$ \Rightarrow x = \dfrac{1}{a} \times \left( {\dfrac{{\mathop a\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}} \right)$
$ \Rightarrow x = \dfrac{a}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$
Thus, we got the value of another variable also.
Hence, the required solution for the system of equations is $x = \dfrac{a}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$ and $y = \dfrac{b}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }}$.

Note: We can also solve the above system of equations using the substitution method or the cross-multiplication method. We can also choose an equation $(ii)$ as $bx + ay = \dfrac{{\mathop {(a + b)}\nolimits^2 }}{{\mathop a\nolimits^2 + \mathop b\nolimits^2 }} - 1$ in the above system of equations and get the same result. If the number of variables is equal to the number of equations, then only the equations will have a solution.