the sum of two consecutive integers is three times their difference.what is the larger number.please help!
2 answers:
Answer:
2
Step-by-step explanation:
The larger number r={2}
PREMISES
r+s=3(r-s)
ASSUMPTIONS
Let r=s+1
Let s=s
CALCULATIONS
(s+1)+s=3[(s+1)-s]
2s+1=3[(s-s)+1]
2s+1=3(0+1)
2s+1=3(1)
2s+1=3
2s+(1–1)=3–1
2s+0=3–1
2s=2
2s/2=2/2
s=2/2
s=
{1}
And if s=1, then
r=s+1=
{2}
PROOF
If r, s={2,1}, then the equations
(1) r+s=3(r-s)
(2) 2+1=3(2–1)
(3) 2+1=3(1) and
(4) 3=3 prove the roots (zeroes) r, s={2,1} of the statement r+s=3(r-s)
HOPE THIS HELPS! :]
9514 1404 393
Answer:
2
Step-by-step explanation:
If n is the larger integer, n-1 is the smaller. The difference of consecutive integers is 1, so their sum is ...
n +(n -1) = 3(1)
2n = 4 . . . . . . . . . add 1
n = 2 . . . . . . . . . . divide by 2
The larger integer is 2.
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Step-by-step explanation:
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