Question

find two consecutive even intergers such that the larger, added to eight times the smaller, equals 110

Answer 1

x,x+1 are your integers. use this to solve it

Answer 2

, let x be the first integer, and x+1 the second

Answer 3

so (x+1)+8x=110, can you take it from there?

Answer 4

x+1+8x=110
9x=109
x=109/9... I think

Answer 5

can u show me step bu step @fall2012

Answer 6

weird... that x is not an integer

Answer 7

sorry.. haha. x and x+2 are u nbers

Answer 8

x+2+8x=110
9x=108
x=12

Answer 9

were do u get the 9 from

Answer 10

x+8x=9x

Answer 11

sure. since the integers are consecutive even integers (thanks for picking that out @josiahh), it means that the larger integer would be the smaller integer plus two

Answer 12

so your answer should be 12 and 14... thx @fall2012

Answer 13

how did u get 108

Answer 14

110-2=108

Answer 15

but theres no - sign in the equation

Answer 16

since the 8 times the larger integer (x+2) is equal to 110, we have
(x+2)+8x=110

Answer 17

yea but you have to subtract 2 from both sides

Answer 18

so its 12 nd 14

Answer 19

expanding the bracket, we have x+2+8x=110,
then we collect like terms, so we have
x+8x=110-2
9x=108,
x=108/9
x=12
and the second integer will be 12+2,=14

Answer 20

thanks u explained it better to my capacities