If 21*2*1= x find 31*x*2*1 = y/21.
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Answered by
5
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).
Answered by
5
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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