Question 462112
Find three consecutive odd integers such that four times the middle integer is equal to two less than the sum of the first and third integers.
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This is a great problem to practice translating English sentences into mathematical sentences and algebraic expressions.
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Let
a = the first integer
b = the middle integer
c = the middle integer
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We can write "four times the middle integer" as 4b.
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"The sum of the first and third integers" may be written as a + c.
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"Two less than the sum of the first two integers" may be written as (a + c) - 2.
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Putting it all together, "four times the middle integer is equal to two less than the sum of the first and third integers," may be written as the equation:
4b = (a + c) - 2
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Now we have a single equation, but three variables. We will use the fact that the integers are consecutive, odd integers to write expressions for b and c in terms of a.
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The integer b is in the middle so it is two more than a, so
b = a + 2
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The integer c is the largest so it is two more than a.
c = a + 4
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Now we take our equation and substitute a + 1 for b and a + 2 for c:
4b = (a + c) - 2
4(a + 2) = (a + (a + 4)) -2
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Now combine like terms and solve for a.
4a + 8 = a + a + 4 - 2
4a + 8  = 2a + 2
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Subtract 2a from both sides, and simplify.
4a + 8 - 2a = 2a + 2 - 2a
2a + 8 = 2
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Subtract 8 from both sides, and simplify.
2a = -6
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Divide both sides by 2.
a = -3
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Since a, b, and c are consecutive intergers with a the smallest, the numbers are -3, -1, and 1.
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Let's check our work using the original English sentences. "Four times the middle integer is equal to two less than the sum of the first and third integers." 
4 times -1 is -4. 
The sum of -3 and 1 is -2. 
-4 is 2 less than -2. (TRUE!)
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Hope this helps!
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Ms.Figgy
math.in.the.vortex