Question

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If ax + by = a2 – b2 and bx + ay = 0, find the value of (x + y).

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Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equation is

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

and

$\rm :\longmapsto\:bx + ay = 0 - - - - (2)$

On adding equation (1) and equation (2), we get

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

$\rm :\longmapsto\:ax + bx + ay + by= (a + b)(a - b)$

$\rm :\longmapsto\:(a+ b)x + (a + b)y= (a + b)(a - b)$

$\rm :\longmapsto\:(a+ b)(x +y)= (a + b)(a - b)$

$\bf\implies \:x + y = a - b$

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ADDITIONAL INFORMATION

Let us consider two linear equations

$\tt \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0$

then

(1). System of equations have unique solution iff

$\bf \:\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$

(2). System of equations have infinitely many solutions iff

$\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

(3). System of equations have no solution iff

$\bf\:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given pair of linear equation is

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

and

$\rm :\longmapsto\:bx + ay = 0 - - - - (2)$

On adding equation (1) and equation (2), we get

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

$\rm :\longmapsto\:ax + bx + ay + by= (a + b)(a - b)$

$\rm :\longmapsto\:(a+ b)x + (a + b)y= (a + b)(a - b)$

$\rm :\longmapsto\:(a+ b)(x +y)= (a + b)(a - b)$

$\bf\implies \:x + y = a - b$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

Let us consider two linear equations

$\tt \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0$

then

(1). System of equations have unique solution iff

$\bf \:\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$

(2). System of equations have infinitely many solutions iff

$\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

(3). System of equations have no solution iff

$\bf\:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$