write three consecutive integers preceding - 11.
Answers
Answer:
*
Step-by-step explanation:
Hi Pam,
Suppose the middle of the three consecutive integers is x then the three numbers are x - 1, x, and x + 1. Thus the greatest is x + 1 and the least is x - 1.
Can you complete the problem now?
If you need more help write back,
Penny
Pam wrote back and received two responses
I know how to figure consecutive intergers ex: x + (x+1) + (x+2)= a given number, but where I get lost on this question is "such that the greatest increased by twice the least is 293". .This just doesn't turn the light bulb on in my head in how to set up the equation.
I know if I could just figure out the set up I would be able to solve it & then feel silly that I didn't see it before.
Pam,
If the three numbers are x, x+1 and x + 2 as you stated then the greatest is x + 2 and the least is x. Thus
the greatest increased by twice the least
is
x + 2 increased by twice x
That is
x + 2 increased by 2x
which is
x + 2 + 2x
Can you finish it now?
Penny
Second Response
Before solving a problem, you have to understand it. A good way to understand this kind of problem is to experiment with actual numbers:
"three consecutive integers such that the greatest increased by twice the least"
Fine! take 3, 4, and 5. These are three consecutive integers. The greatest is 5; twice the least is 2*3 = 6; so the greatest increased by twice the least is 11. That's way smaller than the desired number, namely 293, but at least we know what we're doing here.
Now you try 100, 101, and 102.
The greatest is ___.
The least is ___.
Twice the least is ____.
The greatest increased by twice the least is ____.
(Did you get 302?)
We're already pretty close to 293. You should be able to guess the answer from here. But you should also be able to do it without guessing by starting with x, x+1, and x+2.
The greatest is ____.
The least is ____.
Twice the least is ____.
The greatest increased by twice the least is ____.
So what does x have to be for you to end up with 293?
Chris Fisher
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