Question

the sum of two consecutive integers is three times their difference.what is the larger number.please help!​

Answer 1

2

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! :]

Answer 2

9514 1404 393

Answer:

2

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.