Question

Two times the least of three consecutive odd integers exceeds three times the greatest by 15. How do i set it up to solve?

Answer 1

if they are consecutive odd integers they differ by 2, you can call them
\[x,x+2,x+4\]

Answer 2

then two times the first one is \(2x\) which exceeds 3 times the last one \(3(x+4)\) by \(16\)

Answer 3

oops i meant by 15 not 16

Answer 4

that means
\[2x-15=3(x+4)\] or you could write
\[2x=3(x+4)+15\]

Answer 5

lets go with the first one
\[2x-15=3(x+4)\]

Answer 6

Ok thanks a million! @satellite73

Answer 7

multiply out and get
\[2x-15=3x+12\] then solve for \(x\)

Answer 8

yw

Answer 9

so x=-27
@satellite73

Answer 10

that is what i get, yes
we can check it if you like

Answer 11

yes plz

Answer 12

i got -69=-69

Answer 13

they will be \(-27,-25,-23\) and
\[2\times -27=-54\]
\[3\times -23=-69\] and \(-54\) exceeds \(-69\) by \(15\) since \(-69+15=-54\)

Answer 14

looks like it is right

Answer 15

ok thanks

Answer 16

yw