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The greater of two consecutive integers is 15 more than twice the smaller. Find the integers.

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2 Answers

x = integer
x+ 1 = next (consecutive) integer
 
The greater integer (x+1) is (=) twice the smaller (2x)
 
x + 1 = 2x + 15
 
Put x's together and terms together on opposite sides
 
1 + x - x = 2x + 15 - x
1 = x + 15
-14 = x
x+ 1 = -13.
 
Check x + 1 = 2x + 15
substitute for x+1 and X
 
-13 = 2(-14) + 15
-13 = -28 + 15
-13 = -13 Correct on both sides.  There are your two answers.  I substituted -14 for X and -13 for x+1.
 
 
This is an algebra word problem that can be solved by setting up a system of equations. Stay with me! It only sounds complicated, but we will make it easy.
 
We have two consecutive integers, let's call them x and y.
Let's say y comes after x, so that means y = x+1    (consecutive numbers are in sequence, like counting, each number is 1 more than the number before it)
 
Now let's use our other clue from the problem. The larger number is 15 more than twice the smaller.
y is the larger number, and that is 15 more than 2 times the smaller number (x)
y = 15 + 2x
 
Ok, now let's use substitution.
y = x+1, so we plug that into our equation.
 
x + 1 = 15 + 2x
 
Get the variable terms on the left and the constant terms on the right.
x + 1 = 15 + 2x
- 2x             -2x
 
-x + 1 = 15
    - 1        -1
 
-x    =  14
*-1       *-1
 
x = -14
 
Plug our value of x back into our definition of y.
y = -14 + 1
y = -13
 
So we have -14 and -13. This makes sense because integers can be negative, -14 and -13 are consecutive, and if we multiply the smaller number (-14) by 2, then add 15, we get -13. Our answer checks out!