Question

Solve Simultaneous equations.

$\rm 2^{x-1} + 3^{y-1} = 5$
$\rm 2^{x+3} - 3^y = 23$​

Answer 1

Solution :-

→ 2^(x - 1) + 3^(y - 1) = 5

→ (2^x*2^(-1) + (3^y*3^(-1)) = 5

→ 2^x/2 + 3^y/3 = 5

Splitting RHS part we get,

→ (2^x)/2 + (3^y)/3 = 2 + 3

→ (2^x)/2 + (3^y)/3 = (4/2) + (9/3)

→ (2^x)/2 + (3^y)/3 = (2²/2) + (3²/3)

Comparing LHS = RHS , we get,

→ x = 2

→ y = 2

___________________

Now, Either we put this value in second Equation and check if it satisfy or we can solve second Eqn. similarly,

Lets Try :-

→ 2^(x + 3) - 3^y = 23

→ 2^x * 2^3 - 3^y = 23

→ 2^x * 8 - 3^y = 23

→ 8*2^x - 3^y = 23

Splitting RHS part Now,

→ 8*2^x - 3^y = 32 - 9

→ 8*2^x - 3^y = 8*4 - 9

→ 8*2^x - 3^y = 8*2² - 3²

Comparing LHS = RHS , we get,

→ x = 2

→ y = 2

________

Hence, we can conclude that, value of x and y will be 2 .

Note :- i Tried Assumption method here as values were small . Solution will be difficult for higher values.

Answer 2

Answer:

HOPE IT HELPS.............