Find three consecutive integers such that twice the greatest integer is 2 less than 3 times the least integer.
1 answer:
Givens
Let the first integer = x
Let the second integer = x + 1
Let the third integer = x + 2
Equations
2(x + 2) = 3x - 2 Remove the brackets
Solve
2x + 4 = 3x - 2 Subtract 2x from both sides.
4 = 3x - 2x - 2
4 = x - 2 Add 2 to both sides
4 + 2 = x
x = 6
Answer
The smallest integer is 6
The middle one is 7
The largest one is 8
Check
2*8 = 16
3*6 - 2 = 16 They check.
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