Question

Pls solve on paper using eliminate method​

Answer 1

Answer:

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Answer 2

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

Given pair of equations is

$\rm :\longmapsto\:\dfrac{x}{a} + \dfrac{y}{b} = 2 - - - - (1)$

and

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

Consider,

$\rm :\longmapsto\:\dfrac{x}{a} + \dfrac{y}{b} = 2 - - - - (1)$

Now, Equation (1) can be rewritten as

$\rm :\longmapsto\:\dfrac{bx + ay}{ab} = 2$

$\rm :\longmapsto\:bx + ay = 2ab - - - - (3)$

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

$\rm :\longmapsto\:abx - {b}^{2} y = b{a}^{2} - {b}^{3} - - - - - (4)$

Now, multiply equation (3) by a, we get

$\rm :\longmapsto\:abx + {a}^{2} y = 2b {a}^{2} - - - - (5)$

Now Subtracting equation (4) from equation (5), we get

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

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

$\bf\implies \:y = b$

On substituting the value of y = b in equation (2), we get

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

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

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

$\bf\implies \:x = a$

Hence,

The solution of linear equations

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

and

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

is

$\underbrace{\boxed{ \tt{ \: x \: = \: a \: \: \: \: \: and \: \: \: \: \: y \: = \: b \: }}}$

Verification :-

Consider equation (1),

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

On substituting the values of x and y we get

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

$\rm :\longmapsto\:1 + 1 = 2$

$\rm :\longmapsto \: 2= 2$

Hence, Verified

Consider equation (2),

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

On substituting the values of a and b, we get

$\rm :\longmapsto\:a(a) - b(b) = {a}^{2} - {b}^{2}$

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

Hence, Verified