Question 148112
Hi, Hope I can help,
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The Product of two consecutive positive integers is 1 more than their sum. Find the integers.
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The first integer is "x", the second is (x+1)
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consecutive means one after the other, if there was a third, fourth, and fifth integer it would be (x+2),(x+3),(x+4)(You will do this on any problem where it says consecutive(unless it says consecutive odd, or even, then you add "2" each time
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Now we can do the equation
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The Product of two consecutive positive integers is 1 more than their sum. 
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(x)(x+1) = (x)+(x+1)+1
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We will use the distributive property
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{{{ x^2+x = (x)+(x+1)+1 }}}
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{{{ x^2+x = x+x+1+1 }}}
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{{{ x^2+x = 2x+2 }}}
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We will move everything on the left side
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{{{ x^2+x - 2x -2 = 2x -2x+2-2 }}}
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{{{ x^2 - x - 2 = 0 }}}
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{{{ x^2 - x - 2 }}} has factors of (x+1), and (x-2) ( factors of (-2) add up to (-1))
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We can now solve for "x", using one factor at a time.
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First factor
(x+1)=0
x+1=0
x+1-1=0-1
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x=(-1)
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Second factor
(x-2)=0
x-2=0
x-2+2=0+2
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x = 2
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You can check by replacing "x" with (-1), or 2, in the equation
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{{{ x^2 - x - 2 = 0 }}} ( We will only check one answer, (2)
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{{{ 2^2 - 2 - 2 = 0 }}} 
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{{{ 4 - 2 - 2 = 0 }}} 
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{{{ 0 = 0 }}} 
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The first integer is x, or 2(It asked for a positive integer), the second integer is (x+1), or 3
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You can check by replacing "x" with 2 in 
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(x)(x+1) = (x)+(x+1)+1
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(2)(2+1) = (2)+(2+1)+1
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(2)(3) = (2)+(3)+1
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6 = 6
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First integer = 2
Second integer = 3
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Hope I helped, Levi