2. Find three positive consecutive integers such that the product of the first and the third is one less than six times the second.

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so consecutive integers are notated as x,x+1,x+2 since they are that distance frome ach other such that the product (that means mulitipication) of the first and third ( that means x times (x+2)) is ( that means equals) one less than (minus 1) six times the second (6 times (x+1)) x times (x+2)=-1+6(x+1) distribute using distributiver property which is a(b+c)=ab+ac x(x+2)=x^2+2x 6(x+1)=6x+6 so we have x^2+2x=-1+6x+6 add liket erms -1+6x+6=6x+6-1=6x+5 x^2+2x=6x+5 make one side zeroe so we can factor subtract 6x from both sides x^2-4x=5 subtract 5 from both sides x^2-4x-5=0 factor find what 2 number multiply to get -5 and add to get -4 the answer is -5 and +1 so we put them in front of x in the factored form to get (x-5)(x+1)=0 set each to zero x-5=0 x+1=0 solve for x x-5=0 add 5 to both sides x=5 x+1=0 subtract 1 x=-1 false since the question specified positive integers therefor x=5 the numbers are x,x+1,x+2 x+1=5+1=6 x+2=5+2=7 the numbers are 5,6,7

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