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Algebra
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the difference of two numbers is 3 ; and three times the greater exceeds twice the less by 18 . find the numbers
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let the numbers be x and y \[x-y=3\] \[3x=2y+18\] \[x=3+y\] substitute \[3(3+y)=2y+18\] solve for y \[9+3y=2y+18\] \[y=9\] earlier on we found that \[x=3+y\] we can now find x \[x=3+9=12\]
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