Question

find the two consecutive odd integers such that 1/3 the smaller plus the larger equals 7 more than the sum of the two numbers

Answer 1

Answer:

x=-9

x+2=-7

Step-by-step explanation:

Let

x and (x+2)= the two consecutive integers

x=smaller odd integer

x+2=larger odd integer

1/3 of the smaller integer plus the larger integer =7 more than the sum of the two numbers

1/3x+(x+2)= (x)+(x+2)+(7)

x/3+x+2=x+x+2+7

x+3/3=2x+9

3/3x=2x+9

x=2x+9

x-2x=9

-x=9

x= -9

The smaller odd integer=x= -9

The larger odd integer=(x+2)=-9+2

= -7