19. Find 3 consecutive integers such that twice the least integer is 12 more than the greatest
integer.

et's start by defining the three consecutive integers.

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If x represents the least of the three, then the next consecutive integer is x+1 and the

next consecutive integer is 1 more than that ... or x + 2

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So the three consecutive integers are x, x+1, and x+2.

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Twice the least integer is, as you correctly wrote, 2x.

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12 more than the greatest integer is x+2 + 12 = x + 14

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Since these two amounts must be equal, you can write:

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2x = x + 14

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Now you need to get rid of the x on the right side so that you end up with all the terms

containing x on one side of the equation and everything else on the other side. You can

get rid of the x on the right side by subtracting x from the right side. But if you subtract

something from the right side, you must also subtract it from the left side to keep the

equation in balance. So subtract x from both sides. When you do that, the x disappears

from the right side and the left side becomes 2x minus x which is just x.

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So the equation is reduced to:

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x = 14

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If x is 14 then the next two consecutive integers are 15 and 16 ... x+1 and x+2. So the

three consecutive integers are 14, 15, and 16.

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Check ... the least integer is 14. Twice the least integer is 28. Is 28 actually 12 more

than the greatest integer which is 16? Yes it is, so the problem checks.

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You were on the right track, but you just needed to know the difference in the three integers

was that they were x, x+1, and x+2. Hope this helps to show you how you could work the

problem through to get the final answer.

Step-by-step explanation:

i'm not entirely sure

step-by-step explanation:

if you ask home depot or lowe's they might be able to you out

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step-by-step explanation: