Question

ax+by=1
bx+ay=(a+b)²/a²+b²-1​

Answer 1

Given Question :-

Solve the pair of linear equations :-

$\rm :\longmapsto\:ax + by = 1$

and

$\rm :\longmapsto\:bx + ay = \dfrac{ {(a + b)}^{2} }{ {a}^{2} + {b}^{2} } - 1 - - - (2)$

Answer :-

Basic Concept Used :-

There are 4 methods to solve this type of pair of linear equations.

• 1. Method of Substitution

• 2. Method of Eliminations

• 3. Method of Cross Multiplication

• 4. Graphical Method

We prefer here Method of Eliminations :-

To solve systems using elimination, follow this procedure:

• Step 1: Multiply each equation by a suitable number so that the two equations have the same leading coefficient.

• Step 2: Subtract the second equation from the first so that Equation reduces to one variable.

• Step 3: Solve this new equation for one variable.

• Step 4: Substitute this variable in any of the Equation 1 or Equation 2 above and solve for other variable.

Let's solve the problem now!!

$\rm :\longmapsto\:ax + by = 1 - - - (1)$

and

$\rm :\longmapsto\:bx + ay = \dfrac{ {(a + b)}^{2} }{ {a}^{2} + {b}^{2} } - 1$

can be further simplified as

$\rm :\longmapsto\:bx + ay = \dfrac{ {a}^{2} + {b}^{2} + 2ab - {a}^{2} - {b}^{2}}{ {a}^{2} + {b}^{2} }$

$\rm :\longmapsto\:bx + ay = \dfrac{2ab }{ {a}^{2} + {b}^{2} } - - - (2)$

Now, multiply equation (1) by b, we get

$\rm :\longmapsto\:abx + {b}^{2} y = b - - - (3)$

Now, multiply equation (2) by a, we get

$\rm :\longmapsto\:abx + {a}^{2}y = \dfrac{2b {a}^{2} }{ {a}^{2} + {b}^{2} } - - - (4)$

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

$\rm :\longmapsto\: {b}^{2}y - {a}^{2}y = b - \dfrac{2b {a}^{2} }{ {a}^{2} + {b}^{2} }$

$\rm :\longmapsto\:y( {b}^{2} - {a}^{2}) = \dfrac{ {ba}^{2} + {b }^{3} - {2ba}^{2} }{ {a}^{2} + {b}^{2} }$

$\rm :\longmapsto\:y( {b}^{2} - {a}^{2}) = \dfrac{{b }^{3} - {ba}^{2} }{ {a}^{2} + {b}^{2} }$

$\rm :\longmapsto\:y \cancel{({b}^{2} - {a}^{2})} = \dfrac{b \: \: \cancel{({b }^{2} - {a}^{2})}}{ {a}^{2} + {b}^{2} }$

$\bf\implies \:y = \dfrac{b}{ {b}^{2} + {a}^{2} } - - - (5)$

Now, Substitute the value of 'y' in equation (1), we get

$\rm :\longmapsto\:ax + \dfrac{{b}^{2} }{ {a}^{2} + {b}^{2}} = 1$

$\rm :\longmapsto\:ax = 1 - \dfrac{ {b}^{2} }{ {a}^{2} + {b}^{2} }$

$\rm :\longmapsto\:ax = \dfrac{ {b}^{2} + {a}^{2} - {b}^{2}}{ {a}^{2} + {b}^{2}}$

$\rm :\longmapsto\:ax = \dfrac{ {a}^{2} }{ {a}^{2} + {b}^{2} }$

$\bf\implies \:x = \dfrac{a}{ {a}^{2} + {b}^{2} } - - - (6)$

Hence,

$\boxed{ \bf \: x = \dfrac{a}{ {a}^{2} + {b}^{2} }} \: \: \: \: and \: \: \: \: \boxed{ \bf \: y = \dfrac{b}{ {a}^{2} + {b}^{2} }}$