Question

2^x+3 * 2^2x-y * 5^x+y+3 * 6^y+1/6^x+1 * 10^y+3* 15^x

Answer 1

Answer:

Disclaimer: There is error in the Q. In (II) there should have been 2x - 3y = 12

5x + 3y = 9 -----(I)

2x - 3y = 12 ----- (II)

Add (I) and (II)

7x = 21

x = 3

Putting the value of x = 3 in (I) we get

5(3)+3y=9⇒15+3y=9⇒3y=9-15=-6⇒y=-2

Thus, (x, y) = (3, -6).

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