Question

solve for x and y
a x + by is equal to a square
bx + ay is equal to 2a b​

Answer 1

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

Given pair of equations are

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

and

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

Multiply equation (1) by b and (2) by a, we get

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

and

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

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

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

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

$\purple{\rm\implies \:\boxed{\tt{ \: y \: = \: \frac{ {ba}^{2} }{ {a}^{2} - {b}^{2}}}}}$

Now, Again

On multiply equation (1) by a and (2) by b, we get

$\red{\rm :\longmapsto\: {a}^{2}x + aby = {a}^{3} - - - (5)}$

and

$\red{\rm :\longmapsto\: {b}^{2}x + aby = 2{ab}^{2}- - - (6)}$

On Subtracting equation (5) from (6), we get

$\red{\rm :\longmapsto\: {b}^{2}x - {a}^{2}x= {2ab}^{2} - {a}^{3}}$

$\red{\rm :\longmapsto\:({b}^{2}- {a}^{2})x= {2ab}^{2} - {a}^{3}}$

$\red{\rm\implies \:\boxed{\tt{ x = \frac{a( {2b}^{2} - {a}^{2})}{ {b}^{2} - {a}^{2}}}}}$

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Basic Concept Used

The Elimination Method

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

Step 2: Subtract the second equation from the first to get the new equation in one variable.

Step 3: Solve this new equation for reduced variable to get its value.

Step 4: Substitute this value into either Equation 1 or Equation 2 to get the value of other variable.

Answer 2

Given :

ax + by = a²

bx + ay  = 2ab

To Find : solve for x and y

Solution:

ax + by = a²

bx + ay  = 2ab

Adding both

(a + b)x  + (a + b) y = (a² + 2ab)

=> x + y  = (a² + 2ab)/(a + b)

ax + by = a²

bx + ay  = 2ab

on subtraction

x(a - b)  - y(a - b)  = (a² - 2ab)

=> x - y  =  (a² - 2ab)/ (a - b)

x + y  = a(a + 2b)/(a + b)

x - y  = a (a - 2b)/ (a - b)

Adding both

=> 2x  = a(a + 2b)/(a + b) + a (a - 2b)/ (a - b)

=> x = (a/2) (   (a + 2b) (a - b) + (a - 2b)(a + b) ) /(a² - b²)

=> x = (a/2) ( a² + 2ab - ab - 2b² + a² - 2ab + ab - 2b²) /(a² - b²)

=> x = (a/2)( 2a² - 4b²)/ (a² - b²)

=> x = a( a² - 2b²)/(a² - b²)

On subtraction

y = (a/2) (   (a + 2b) (a - b) - (a - 2b)(a + b) ) /(a² - b²)

=> y =  a²b/ (a² - b²)

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