Need Help, Algebra word problems?

Question

Well I can't solve problems like this one:

The greater of two consecutive integers is 10 greater than twice the lesser. Fine the integers.

Here is something I got out of that: 2x+10 which stands for the greater number, (x as in the lesser number). But I don't know what to do from there, heck, I don't even know if that is right.

Answer 1

You are right about: 2x+10=y.

I had to scratch my head a moment before I realized that they are CONSECUTIVE integers. With consecutive integers each number is the previous plus 1.

x+1=y

or,

2x+10=x+1

x= -9 and y= -8

Answer 2

Break down the problem.

The greater of two consecutive integers is 10 greater than twice the lesser. Fine the integers.

"10 greater than twice the lesser"
Subtracting 10 from the greater one, then diving by 2 will equal the lesser one.

Not sure why it says consecutive though. That would indicate one number right after the other.

Yeah 2x+10 would equal the greater number with x being the lesser.


A thought! Could it be a negative integer? Then ten greater would mean closer to zero, than twice the lesser which would be away from zero, so it would be great if it were positive, but less if negative.

Answer 3

You can do it! Look:

Let x = the lesser number
x + 1 = the greater number

Now, follow the words,

x + 1 = 2x + 10

There's your equation (remember, is means =). Let's solve it:

subtract x from both sides:

1 = x + 10,

subtract 10 from both sides:

-9 = x

So, the two integers would be -9 and -8. Let's see if it works with the words.

Is -8 , 10 more than twice -9. Twice -9 is -18, add 10 and you get -8. So it works!

Answer 4

Use x and x+1 for your consecutive integers...
Now the equation is 2x=x+1+10
Solve for x now...
x=11
Now let's verify it...
Your integers are 11 and 12
11 times 2 = 22
and 12 plus 10 = 22