Question

15x + 22y = 5

40x + 55y = 13

Solve for X and Y

( Substitution method)

Answer 1

Given Equation is 15x + 22y = 5   ---- (1)

Given Equation is 40x + 55y = 13  ---- (2)

Equation (1) can be written as,

= > 15x + 22y = 5

= > 15x = 5 - 22y

= > x = (5 - 22y)/15    ------- (3)

On substituting (3) in (2), we get

= > 40x + 55y = 13

= > 40(5 - 22y/15) + 55y = 13

= > 40(5 - 22y) + 825y = 195

= > 200 - 880y + 825y = 195

= > 200 - 55y = 195

= > -5 = -55y

= > y = 1/11.

Substitute y = 1/11 in (3), we get

= > x = (5 - 22y/15)

= > x = [5 - 22(1/11)]/15

= > x = (5 - 2)/15

= > x = 3/15

=  >x = 1/5

Therefore, the value of x = 1/5 and y = 1/11.

Hope this helps!

Answer 2

$\huge{\bold{Hello!!!}}$

15x + 22y = 5 --------( 1 )

40x + 55y = 13--------( 2 )

From eq'n ( 1 ) , we get

15x + 22y = 5

=> 22y = 5 - 15x

=> y = ( 5 - 15x / 22 ) ------- ( 3 )

Substitute eq'n ( 3 ) in eq'n ( 2 )

40x + 55y = 13

=> 40x + 55( 5 - 15x / 22 ) = 13

=> 40x + ( 275 - 825x / 22 ) = 13

=> 880x + 275 - 825x = 286

=> 55x = 286 - 275

=> 55x = 11

=> x = 11 / 55

=> x = 1 / 5 >>> Ans

From eq'n ( 3 )

y = ( 5 - 15x / 22 )

=> y = ( 5 - 15( 1 / 5) / 22)

=> y = 5 - 3 / 22

=> y = 2 / 22

=> y = 1 / 11 >>> Ans


Hence the value of x = 1 / 5 and y = 1 / 11
___________________

$\textbf{Verification}$

15x + 22y = 5 --------( 1 )

40x + 55y = 13--------( 2 )

Eq'n ( 1 )

LHS = 15 ( 1 / 5 ) + 22 ( 1 / 11 )

=> 3 + 2 = 5

= RHS

Eq'n ( 2 )

LHS = 40( 1 / 5 ) + 55 ( 1 / 11 )

=> 8 + 5 = 13

= RHS

$\boxed{LHS = RHS}$

$\textbf{Hence, it is verified}$

$\boxed{Thanks}$

$\textbf{Hope it helps!!!}$