SOLUTION: the sum of 2 consecutive integers is 27 more than 5 times the larger integer find the integers

Question 249475: the sum of 2 consecutive integers is 27 more than 5 times the larger integer find the integers
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
We always seek ways to use the fewest unknowns when solving word problems.
.
We could call the two integers 'x' and 'y'. But we are told they are consecutive, so we can say:
x = x
y = x+1
That gives us only one unknown, 'x'.
.
We also are given some relationships:
the sum of two consecutive integers: x + x+1
is 27 more than 5 times the larger integer: 5(x+1) + 27
.
So,
x + x+1 = 5(x+1) + 27
.
Collecting and simplifying terms
2x + 1 = 5x + 5 + 27
2x + 1 = 5x + 32
.
Subtracting 1 from both sides
2x = 5x + 31
.
Subtracting 5x from both sides
-3x = 31
.
Dividing both sides by -3 shows that 'x' is NOT an integer, so there is no solution...
.
If only the setup had been 26 more instead of 27. (Or is that a typo?)
Then we would have had:
.
2x + 1 = 5x + 5 + 26
2x + 1 = 5x + 31
2x = 5x + 30
-3x = 30
x = -10
.
Then the two consecutive integers would have been: -10 and -9.
-10 + (-9) = -19
.
5*(-9) = -45
-45 + 26 = -19
.
Done.
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