Number problem

[tertie]

If one-half of one ineger is subtracted from three-fifths of the next consecutive integer, the difference is three. What are the two integers?


3/5(x+1) - 1/2x=3

is this right? if it is, can someone help me with the next step?

Edit: TKHunny - Removed color.

 

[tkhunny]

You don't have a clear definition.

x = The integer
x+1 = The next consecutive integer

Your notation is OK, but is looks like it could lead to misunderstanding on other problems. Use more parentheses.

(3/5)*(x+1) - (1/2)*(x) = 3

Next? I'd multiply everything be the least common denominator, 10.

 

[tertie]

I have done that and come up with x=9

This does not make sense because 3/5 of 10 =6 and 1/2 of 9=4.5

6-4.5 does not equal 3

what did i do wrong?

 

[tkhunny]

It's a little tough to provide an opinion if you don't show what you are doing. Multiply that equation by 10 and show us what you get in the ONE STEP. No jumping about. One step at a time.

Good job on checking your answer. Now let's find one that works.

 

[tertie]

ok, here is what i did...

10 times 3/5 (x+1) - 10 times 1/2x=3

6(x+1) -5x=3
6x+6-5x=3
x+6=3
x=9

Ok i think I see what i did wrong. according to this x=-3 (minus 6 from both sides) but even -3 does not work either.

Did I miss a step?

 

[tkhunny]

What happened to 10*3 = 30 on the right hand side?

 

[tertie]

my brain is fried....had a dumb moment

man now I feel really stupid! I am sorry :-(

I got it now x=24

3/5 of 25 = 15
1/2 of 24 = 12

leaves a difference of three!
Woohoo!!

Thank you for your time and help, (and your patience too)

 

[tkhunny]

No worries. That's why I asked you to show that. Learning.

Note to ALL Readers:

See how much more can be accomplished when you SHOW YOUR WORK?!??!?!?!?!

 

[tertie]

Can I ask for assistance one more time?

A plane flies at 720 mi against a steady 30-mi/h headwind and returns to the same point with the wind. If the entire trip takes 10 h, what is the plane's speed in still air?

now I know that d=r(t)

x=720(10) = 7200


now do I divide by 30? (=240) than add that to the 720??

 

[Gene]

You should post new problems in a new post.
When flying against the wind it slows down. The rate is r-30. With the wind it is r+30.
d does = rt so
720=(r-30)t₁ and
720=(r+30)t₂
t₁+t₂ = 10

 

[Mrspi]

Can I ask for assistance one more time?

A plane flies at 720 mi against a steady 30-mi/h headwind and returns to the same point with the wind. If the entire trip takes 10 h, what is the plane's speed in still air?

now I know that d=r(t)

x=720(10) = 7200


now do I divide by 30? (=240) than add that to the 720??

You've already received one suggestion as to how to set up the problem. I'll offer a slightly different approach.

Let r = speed of plane in still air

Flying against the wind, the plane's speed will be reduced by the wind speed, and its rate will be r - 30 mph. We know that distance = rate * time, and that distance/rate = time. So,
time flying against the wind = 720/(r - 30)

Flying with the wind, the plane's speed will be increased by the wind speed, and its rate will be r + 30 mph. So,
time flying with the wind = 720/(r + 30)

We also know this: the total time for the trip was 10 hours. Thus,

time against wind + time with wind = 10

 720        720  
------  + ------- = 10
r - 30     r + 30

Now, you should be able to solve for r. Hint: multiply both sides of the equation by the common denominator for the fractions.

I hope this helps you.