Question

Find two consecutive odd integers such that 1/3 the smaller plus twice the larger equals 7 more than the sum of the two numbers. @Algebra I

Answer 1

See like i know its starts like: x, x+2
x+x+2=1/3x + 2(x+2)........... or something of that nature :P

Answer 2

This is a system of equations.

x and x+2
x/3+2(x+2)+7=x+x+2
x/3+2x+4+7=2x+2
7x/3+11=2x+2
x/3=-9
x=-27
x+2=-25