Find two numbers such that the sum of twice the first and thrice the second is 103 and dour times the first exceeds seven times the second by 11
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Let the numbers be x, y.
According to the question,
2x + 3y = 103 ...eqn(1)
4x = 7y+11
=> 4x - 7y = 11 .... Eqn(2)
Eqn(1) * 2 - eqn(2)
4x + 6y = 206
4x -7y = 11
========
13y = 195
y = 15 .
Substitute in eqn(1)
2x+3(15) = 103
2x + 45 = 103
2x = 103 - 45
2x = 58
x = 29 .
The numbers are 29 , 15
According to the question,
2x + 3y = 103 ...eqn(1)
4x = 7y+11
=> 4x - 7y = 11 .... Eqn(2)
Eqn(1) * 2 - eqn(2)
4x + 6y = 206
4x -7y = 11
========
13y = 195
y = 15 .
Substitute in eqn(1)
2x+3(15) = 103
2x + 45 = 103
2x = 103 - 45
2x = 58
x = 29 .
The numbers are 29 , 15
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