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Algebra 1: Word Problems... Help?
#1
Hey guys, I am really struggling in these Word Problems...

An example would be:

Find four consecutive integers with a sum of 94.

Let x = the first consecutive integer
And x + 2 = the second consecutive integer
And x + 3 = the third consecutive integer
And x + 4 = the fourth consecutive integer

Now they want me to find all the integers and its so weird can somebody explain?... It has to equal 94... Please help! Big Grin
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#2
(10-19-2010, 06:52 PM)L3g1tWa5te Wrote: Hey guys, I am really struggling in these Word Problems...

An example would be:

Find four consecutive integers with a sum of 94.

Let x = the first consecutive integer
And x + 2 = the second consecutive integer
And x + 3 = the third consecutive integer
And x + 4 = the fourth consecutive integer

Now they want me to find all the integers and its so weird can somebody explain?... It has to equal 94... Please help! Big Grin

I believe you made an error in your post, or there's a discrepancy in the problem itself. If the problem wants to you find four consecutive integers then it'd be:

x
x + 1
x + 2
x + 3


By jumping immediately from x to (x + 2) you're breaking consecution. Regardless, you need to write out the equation based on what they've given you. You know that it's the sum of four consecutive integers, represented by x, which each iteration adding one more than the last. Thus the problem can be represented as:

x + (x + 1) + (x + 2) + (x + 3) = 94

And of course you can simplify that equation.

4x + 6 = 94
4x = 88
x = 22

Then check your answer:

x + (x + 1) + (x + 2) + (x + 3)
22 + (22 + 1) + (22 + 2) + (22 + 3)
22 + 23 + 24 + 25 = 94
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How art thou, thou globby bottle of cheap, stinking chip oil?
Come and get one in the yarbles, if ya have any yarbles, you eunuch jelly thou!
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#3
(10-20-2010, 08:25 AM)Disease Wrote: I believe you made an error in your post, or there's a discrepancy in the problem itself. If the problem wants to you find four consecutive integers then it'd be:

x
x + 1
x + 2
x + 3


By jumping immediately from x to (x + 2) you're breaking consecution. Regardless, you need to write out the equation based on what they've given you. You know that it's the sum of four consecutive integers, represented by x, which each iteration adding one more than the last. Thus the problem can be represented as:

x + (x + 1) + (x + 2) + (x + 3) = 94

And of course you can simplify that equation.

4x + 6 = 94
4x = 88
x = 22

Then check your answer:

x + (x + 1) + (x + 2) + (x + 3)
22 + (22 + 1) + (22 + 2) + (22 + 3)
22 + 23 + 24 + 25 = 94


Ha, you aced the problem, and now I get it MUCH better! Thanks so much Disease!
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#4
Great explanation Disease, very detailed Thumbsup
I probably would've just used the guess and check method instead of doing it out, it's easy enough
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#5
(10-20-2010, 06:42 PM)L3g1tWa5te Wrote: Ha, you aced the problem, and now I get it MUCH better! Thanks so much Disease!

Your a Mathermagition Big Grin
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#6
(10-20-2010, 08:25 AM)Disease Wrote: I believe you made an error in your post, or there's a discrepancy in the problem itself. If the problem wants to you find four consecutive integers then it'd be:

x
x + 1
x + 2
x + 3


By jumping immediately from x to (x + 2) you're breaking consecution. Regardless, you need to write out the equation based on what they've given you. You know that it's the sum of four consecutive integers, represented by x, which each iteration adding one more than the last. Thus the problem can be represented as:

x + (x + 1) + (x + 2) + (x + 3) = 94

And of course you can simplify that equation.

4x + 6 = 94
4x = 88
x = 22

Then check your answer:

x + (x + 1) + (x + 2) + (x + 3)
22 + (22 + 1) + (22 + 2) + (22 + 3)
22 + 23 + 24 + 25 = 94
Exactly what I was going to post. I am in algebra 1 also, and we are learning this crap. I got a 98 in it last quarter. too easy imo.
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