Question 146850
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/6 = 14/6 }}}
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 {{{ x = 14/6 }}}
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x = {{{ 14/6 }}}, or {{{ 7/3 }}}
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We can check by replacing "x" with {{{ 7/3 }}}
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 ( 8x = 2(x+2) + 10 )
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 {{{ 8( 7/3) = 2((7/3)+2) + 10 }}}
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 {{{ 56/3 = 2(13/3) + 10 }}}
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 {{{ 56/3 = (26/3) + 10 }}}
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 {{{ 56/3 = 56/3 }}}
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First number = "x" or  {{{ 7/3 }}} ( 2 {{{ 1/3 }}} )
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Second number = (x+2) or {{{ 13/3 }}} ( 4 {{{ 1/3 }}} )
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Third number = (x + 4) or {{{ 19/3 }}} (6 {{{ 1/3 }}} )
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Hope I helped, Levi