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3 years ago

Find two consecutive integers such that the larger is nine more than twice the smaller

1 answer:

7 0

If x is one integer and x+1 is the next, we have 2*x+9=x+1. Subtracting 9 and x from both sides, we get x(the smaller number)=-8 and the larger number is -7.

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mestny [16]

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Answer:

Step-by-step explanation:

Let x represent the number. The problem statement tells you ...

2/3(3x +6) = 10

2x +4 = 10 . . . . . . use the distributive property

x +2 = 5 . . . . . . . . divide by 2

x = 3 . . . . . . . . . . .subtract 3

The number is 3.

<em>Comment on the solution</em>

The above shows an "alternative" solution method. The more usual way this might be done is ...

2(3x +6) = 30 . . . . . multiply by 3 to clear fractions

6x +12 = 30 . . . . . . . eliminate parentheses

6x = 18 . . . . . . . . . . . subtract 12

x = 3 . . . . . . . . . . . . divide by 6

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Nadusha1986 [10]

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