SOLUTION: I can't seem to figure out how to set up this problem.
Find 3 consecutive integers such that the product of the 2 larger integers is 50 more than 5 times the smallest integer.
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-> SOLUTION: I can't seem to figure out how to set up this problem.
Find 3 consecutive integers such that the product of the 2 larger integers is 50 more than 5 times the smallest integer.
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Question 739877: I can't seem to figure out how to set up this problem.
Find 3 consecutive integers such that the product of the 2 larger integers is 50 more than 5 times the smallest integer. (There are 2 sets of 3 consecutive integers).
I know that to find consecutive integers:
1st: x
2nd: x+1
3rd: x+2
But when I read the word problem again, I get stumped on how to properly plug in the right numbers to be able to move forward.
Here's what I think it is:
1st: x
2nd: (x+1)+50
3rd: (x+2)+5x
I will know how to complete the problem if I can get help on how to correctly set up the equation.
Thank you greatly. Found 2 solutions by josgarithmetic, lwsshak3:Answer by josgarithmetic(39620) (Show Source):
You can put this solution on YOUR website! Find 3 consecutive integers such that the product of the 2 larger integers is 50 more than 5 times the smallest integer.
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let x=smallest consecutive integer
(x+1)=next consecutive integer
(x+2)=largest consecutive integer
..
product of the 2 larger integers=(x+1)(x+2)
5 times the smallest integer=5x
adding 50 to the 5 times the smallest integer to set up the equation.
(x+1)(x+2)=50+5x
x^2+3x+2=50+5x
x^2-2x-48=0
(x-8)(x+6)=0
x=8
or
x=-6
..
For positive integers:
smallest consecutive integer=8
next consecutive integer=9
largest consecutive integer=10
..
For negative integers:
smallest consecutive integer=-6
next consecutive integer=-5
largest consecutive integer=-4
(