SOLUTION: what is eight times the first of three consecutive odd integers is ten more than twice the second

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Question 146850: what is eight times the first of three consecutive odd integers is ten more than twice the second
Answer by Electrified_Levi(103) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help,
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what is eight times the first of three consecutive odd integers is ten more than twice the second
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When it says consecutive odd, it means you add two to the number, then add two more to the first number(x, x+2, x+4,x+6,x+8, and so on)(you also do it to consecutive even, if it only says consecutive integers then you only add 1 every time)
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Our three consecutive odd integers are (x, x+2, x+4)
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Our equation is
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what is eight times the first of three consecutive odd integers is ten more than twice the second
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( 8x = 2(x+2) + 10 )
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We will use distribution on the right side
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( 8x = 2x + 4 + 10 )
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We will add the right side
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( 8x = 2x + 14 )
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We will move "2x to the left
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( 8x - 2x = 2x - 2x + 14 )
( 6x = 14 )
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We will divide everything by "6"
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+6x%2F6+=+14%2F6+
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+x+=+14%2F6+
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x = +14%2F6+, or +7%2F3+
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We can check by replacing "x" with +7%2F3+
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( 8x = 2(x+2) + 10 )
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+8%28+7%2F3%29+=+2%28%287%2F3%29%2B2%29+%2B+10+
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+56%2F3+=+2%2813%2F3%29+%2B+10+
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+56%2F3+=+%2826%2F3%29+%2B+10+
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+56%2F3+=+56%2F3+
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First number = "x" or +7%2F3+ ( 2 +1%2F3+ )
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Second number = (x+2) or +13%2F3+ ( 4 +1%2F3+ )
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Third number = (x + 4) or +19%2F3+ (6 +1%2F3+ )
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Hope I helped, Levi