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# word problem

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

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

In word problem always choose a variable for unknown( what problem is asking you to find out), and then convert the language of the problem into algebraic expression, and set the relation(equation) and solve.
n : lowest number.
n+1 : 2nd number
n+2 : 3rd number
then set the algebraic equation accordingly, and solve . the key is to choosing a variable and developing
algebraic expression in terms of variable selected and then solve it.
n + (n+1)/2 + 4( n+2) = -30

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.

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

Add it up, and YES! it checks.