find three consecutive numbers such that the sum of the first integer, half the second integer, and four times the third integer is -30

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# 2 Answers

n + (n+1)/2 + 4( n+2) = -30

11n/2 + 17/2 = -30

11n/2 = -30 -8.5

11n = -77

n = -7

-7, -6, -5

You have to begin this kind of problem by defining your unknowns. Since there are three unknowns, but we know they are all consecutive integers, let's call them x, x+1 and x+2.

Now let's put in the operations that we see in the problem:

They are all summed..... this means we're going to add them up.

We have to cut the second one in two....... that is, divide it.

And we have to multiply the last one by 4.

Finally we have to set the sum equal to negative 30.

so, that gives us the number sentence that Parviz started with above.

x + (x+1)/2 + 4(x+2) = -30

To begin to solve this for x (our smallest integer). I would first mulitply through the entire equation (both sides) by 2.

That gives 2x + (x+1) + 8(x+2) = -60

Distribute the 8 and collect like terms on the left to get

11X + 17 = -60

then subtract 17 from both sides:

11X = -77

divide by 11 on both sides and get the solution, x = -7

so our other integers are one more and two more than this, or -6 and -5.

It's always a good idea to check your answers.

Does -7 +(-7+1)/2 + 4(-7+2) = -30 ?

Add it up, and YES! it checks.

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