You can put this solution on YOUR website! find two numbers
a & b
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such that the sum of twice the first and thrice the second is 92
2a + 3b = 92
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and four times the first exceeds 7 times the second by 2
4a = 7b + 2
4a - 7b = 2
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Multiply the 1st equation by 2, subtract from the above equation
4a - 7b = 2
4a + 6b = 184
---------------subtraction eliminates a, find b
0 - 13b = -182
b =
b = +14 is the 2nd number
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Find the 1st number using the 1st original equation
2a + 3(14) = 92
2a + 42 = 92
2a = 92 - 42
2a = 50
a = 50/2
a = 25 is the 1st number
:
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Check solutions in the 2nd original equation
4(25) = 7(14) + 2
100 = 98 + 2; confirms our solutions of a=25 and b=14
