Linear Equations-algebraic method-eleventh grade-2

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When A and B sat down to play, B had two-thirds as much money as A. After a while,
A won $ 15 and then he had twice as much money as B. How much money did each have at first?

Solution: Suppose A had x dollars at first

Then B had 2x /3 dollars

A won $ 15. Then A had $ (x + 15)

B had $ (2x/3 - 15)

Therefore, x + 15 = 2(2x/3 - 15)

i.e. x + 15 = 4x/3 – 30

Therefore, 4x/3 – x = 30 + 15

i.e. x/3 = 45

Therefore, x = 3 X 45 = 135

Therefore, A had $ 135 at first

B had $ 2x/3 = (2 X 135)/3 = 90

B had $ 90 at first.

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